unit 8: reactions-key regents chemistry ’14 mr. · pdf fileunit 8: reactions-key regents...

Unit 8: Reactions-Key Regents Chemistry ’14-‘15 Mr. Murdoch

Website upload 2015 of 57 Lecture Key

Unit 8:

Reactions

1.

Student Name: _____________Key_______________

Class Period: _3, 5, & 10_

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Unit 8 Vocabulary:

1. Coefficient: A number placed in front of a formula to balance a

chemical reaction.

2. Decomposition: A redox reaction in which a compound breaks up to

form two elements.

3. Double Replacement: A solution reaction in which the positive ion of

one compound combines with the negative ion of the other

compound to form a precipitate, and the other ions remain dissolved

in solution.

4. Law of Conservation of Charge: Charge may not be created or

destroyed by physical or chemical change. This is the basis for

writing chemical formulas and half-reactions, and for balancing

redox reactions.

5. Law of Conservation of Energy: Energy may not be created or


calculating the heat of reaction.

6. Law of Conservation of Mass: Matter may not be created or


balancing chemical reactions.

7. Mole Ratio: The whole-number ratio between components of a

balanced chemical reaction.

8. Oxidation: The loss of electron(s), causing the oxidation number of a

species to become more positive.

9. Precipitate: An insoluble solid that is formed in either a double-

replacement reaction or as an excess solute added to a saturated

solution.

10. Product: The substances that are formed by a chemical reaction,

shown on the right side of a chemical equation.

11. Reactant: The substances that are the beginning of a chemical

reaction, shown on the left side of a chemical equation.

12. Redox reaction: A reaction in which one element is oxidized and

another element is reduced.

13. Reduction: The gain of electron(s), causing the oxidation number of

a species to become more negative.



14. Single Replacement: A redox reaction in which an element replaces

an ion within a compound.

15. Spectator Ion: An ion that does not participate in the overall

chemical reaction. In a redox reaction, the spectator ion is the ion

whose charge does not change. In a double replacement reaction,

they are the ions that remain dissolved in solution.

16. Stoichiometry: The mathematics of mole relationships.

17. Synthesis: A redox reaction in which two elements combine to form

a compound.



Unit 8 Homework Assignments:

Assignment: Date: Due:



Notes page:



Chemical Equation: A chemical equation is a symbolic representation of

a chemical reaction. The chemical equation includes:

1. the substances starting the process (reactants);

2. the substances being formed during the process (products);

3. the phases of matter for the substances during the process;

4. the number of moles of each substance in the process;

5. and the resulting change in energy of the entire process.

Reactants Products

Coefficients are whole numbers placed in front of the substance symbols

to denote a mole ratio that functions within the Law of Conservation of

Mass.

HCl(aq) + Zn(s) H2(g) + ZnCl2(aq) (reactants) (products)

This equation states that the reactant aqueous hydrochloric acid (HCl)

when combined (+) with solid zinc metal (Zn) undergoes a reaction ()

to produce hydrogen gas (H2) and aqueous zinc chloride (ZnCl2).

Topic: Chemical Equations

Objective: What is a chemical equation?



Conservation of Mass:

The Law of Conservation of Mass states that mass may not be created

or destroyed by physical or chemical changes.

Any elements found on the reactants (left) side MUST be found on

the products (right) side, and the numbers of moles of those elements

MUST be equal on both sides.

Look at the first equation we used in this unit:

HCl(aq) + Zn(s) H2(g) + ZnCl2(aq)

(reactants) (products)

This equation breaks the Law of Conservation of Mass. Note that there

are unequal numbers of moles of H and Cl on both sides of the equation.

H + Cl(aq) + Zn(s) H (g) + Zn Cl (aq)

We can’t just add matter, so we need to place coefficients to act as

multipliers in front of a substance’s formula:

HCl(aq) + Zn(s) H2(g) + ZnCl2(aq)

We didn’t have to change anything with the zinc (Zn), as there is only

one mole (the 1 is not shown) of Zn on either side.

This states that 2 moles of HCl are required to react with 1 mole of Zn.

Watch YouTube Conservation of Mass video (1:25)

https://www.youtube.com/watch?v=mcnga-bbNXk

Topic: Law of Conservation of Mass

Objective: How is the Conservation of Mass applied to equations?





Writing Chemical Equations:

If you are given the names in your equation, the steps for writing a

chemical equation are as follows:

1. Balance the Equation:

i. Write in PENCIL (you’ll make mistakes; we all do!);

ii. Write the coefficients in one element at a time;

iii. You may only add coefficients; you may NOT change any

chemical formulas when balancing the equation;

iv. Revise (remember, to use PENCIL) the coefficients where

necessary.

*Note: A coefficient is the number in front; a subscript is behind!

When you write 2 Cl, that states there are TWO atoms of chlorine.

When you write Cl2, that states there is ONE molecule of diatomic

(2 atoms) chlorine. Diatomic molecules of (Br2, I2, N2, Cl2, H2, O2,

& F2) exist whenever these elements are not in a compound with

another element.

In NaCl, there is one Cl-1

ion (due to Na’s+1

charge), but when Cl-1

is

separated from another compound, the two single Cl-1

ions covalently

share their combined 14 valence electrons as one diatomic Cl2 molecule.

Topic: Rules for Writing Equations

Objective: What are the rules for writing chemical equations?



NaCl Na + Cl

Again, the separated Cl-1

ions will form a diatomic molecule, which

“adds” matter:

NaCl Na + Cl2

However, this throws off the coefficient balance.



Balancing an Equation:

Look at the equation of NaCl Na + Cl2:

i. There is one Na on the left and one Na on the right. Na is

balanced…for the moment.

ii. There is one Cl atom on the left and two Cl atoms on the right.

Multiply them together (1 x 2 = 2):

This means when balanced, there should be two chlorine

atoms on BOTH sides of the arrow.

iii. As there exists 2 Cl atoms on the right (products) side, place a

coefficient of in front of the entire NaCl formula on the left side:

That makes NaCl Na + Cl2

iv. However, now look at Na on the right (products) side of the arrow.

When we added the 2 coefficient in front of NaCl, we not only

multiplied the Cl by 2, but we multiplied the Na by 2 as well.

v. So, add a coefficient of in front of the Na on the right (products)

side:

That gives us 2 NaCl Na + Cl2

vi. Doing the math, there are 2 Na atoms & 2 Cl atoms on the left

(reactants) side, and 2 Na atoms & 2 Cl atoms on the right

(products) side. The equation (and reaction) is BALANCED!!!

Watch YouTube Bozeman Science Balancing Equations video (9:38)

https://www.youtube.com/watch?v=wxvERNlUdBQ

Topic: Balancing Equations

Objective: How do we balance the coefficients used in an equation?





Balancing an Equation example #1:

H2 + O2 H2O

i. There are two H’s on the left and two H’s on the right. OK…for

now.

ii. There are two O’s on the left, and only one O on the right. Multiply

(2 x 1 = 2) and place the resulting coefficient of in front of the

H2O on the right:

H2 + O2 H2O

There will now be two O’s on either side of the arrow, but

now there are four total H’s on the right.

iii. Solve this by placing a coefficient of in front of the H2 on the left:

H2 + O2 2 H2O

There are now four H’s on BOTH sides, and two O’s on

BOTH sides.

iv. The equation is balanced!




Al + O2 Al2O3

i. There is one Al atom on the left and two Al atoms on the right.

Multiply (1 x 2 = 2) and place the resulting coefficient of in front

of the Al on the left:

Al + O2 Al2O3

ii. There will now be two O’s on the left and three O’s on the right.

Multiply (2 x 3 = 6) which means there should be six O’s on both

sides.

iii. On the left side of the arrow there are two O’s. So, (2 x Y = 6), with

Y = 3. Therefore place a in front of the O2 on the left.

2 Al + O2 2 Al2O3

iv. On the right side of the arrow there are three O’s (in Al2O3). Divide

6 by 3 and you get 2, so place a on the right side:

2 Al + 3 O2 Al2O3

Uh-oh. Al is now wrong, as there are only two Al atoms on

the left, and a total of four Al atoms on the left.

v. (you used PENCIL, right?) the coefficient of on the

left side with a coefficient of :

Al + 3 O2 2 Al2O3

There are now four Al’s and six O’s on either side of the

arrow.

vi. Balanced!




SO3 SO2 + O2

i. Note here that O is found in THREE places; once on the left, twice

on the right.

ii. There is one S on both sides. S is balanced…for a bit.

iii. There are three O’s on the left side of the arrow, but a total of four

O’s on the right side of the arrow. Multiply (3 x 4 = 12) which

means there should be 12 atoms of O on either side when balanced.

Divide the three O’s on the left by 12, which leaves 4, so place a

coefficient of in front of the SO3:

SO3 SO2 + O2

We now have 12 total O’s and 4 total S’s on the left.

iv. That means we have to add a coefficient of in front of the right

side SO2 to have 4 S’s on the right.

4 SO3 SO2 + O2

This gives 8 total O’s in the SO2, but we need 12 total O’s on

the right.

v. So, (12 – 8 = 4), so place a coefficient of in front of the O2 (2 x

O2 = 4 O) to give a total of 12 O’s on the right.

4 SO3 4 SO2 + O2

vi. Balanced! But look at all the coefficients. They have a common

factor, in this case 2. If you can simplify ALL the coefficients, do

so! In this case, divide ALL coefficients by 2.

SO3 SO2 + O2 (the 1 does NOT need to be given)




Ca + HNO3 Ca(NO3)2 + H2

i. First thing…treat the ENTIRE polyatomic ion (NO3) as one

element. That means any coefficient affects ALL of the polyatomic

ion atoms. (It may be easier to place any polyatomic ions in

parentheses during balancing.)

Ca + H(NO3) Ca(NO3)2 + H2

For now, there is one Ca on both sides of the arrow.

ii. There is one H on the left, and two H on the right. So, (1 x 2 = 2),

so there should be 2 H’s on both sides. Place a coefficient of in

front of HNO3:

Ca + H(NO3) Ca(NO3)2 + H2

That gives two (NO3) polyatomic ions on both sides, one Ca

on both sides, and two H’s on both sides.

iii. Balanced!




FeCl3 + Pb(NO3)2 Fe(NO3)3 + PbCl2

i. Whoa! There is a LOT here to work with! There are FOUR things

to balance.

ii. Take one species at a time; be patient and do one action at a time.

iii. Start simple; there is one Fe on each side; let that be…for now.

iv. Look at the lead; again, there is only one Pb on each side, so OK.

v. There are three Cl atoms on the reactant side, and two Cl atoms on

the product side. Multiply (2 x 3 = 6), so we need six Cl’s on either

side.

vi. Look at the reactants; divide 6 by the three Cl’s we have on the left,

getting 2, so our coefficient for the reactant FeCl3 will be .

FeCl3 + Pb(NO3)2 Fe(NO3)3 + PbCl2

vii. Look at the products; divide 6 by the two Cl’s we have on the left,

getting 3, so our coefficient for the product PbCl2 will be .

2 FeCl3 + Pb(NO3)2 Fe(NO3)3 + PbCl2

viii. Now let’s work with the nitrate polyatomic ion. There are two

NO3’s in the reactants and three NO3’s in the products. Again, it is

(2 x 3 = 6), so we’ll need six nitrate ions on either side of the arrow.

ix. Place a coefficient of for the reactant with nitrate, and a coefficient

of for the product that contains the nitrate ion.

2 FeCl3 + Pb(NO3)2 Fe(NO3)3 + 3 PbCl2

x. Hey! Look at the metals! Fe and Pb BOTH have the same amounts

on either side! Note that this doesn’t often happen, but when it

does, sweet!

xi. Balanced!



Writing Equations when given the names of the formulas:

If you are given the names of the compounds used in a chemical

reaction rather than their formulas, use can use the same rules for

writing formulas that you used in Unit 7b.

After writing the formulas and setting up the basic equation, balance

the equation normally.

Given a reaction as names, write the formulas:

1. Magnesium metal + lead (IV) nitrate magnesium nitrate + lead metal

Mg(s) + Pb(NO3)4(aq) Mg(NO3)2(aq) + Pb(s)

i. Now balance the raw equation:

Mg(s) + Pb(NO3)4(aq) Mg(NO3)2(aq) + Pb(s)

ii. THIS one was simple!

2. Copper metal + Oxygen gas copper (I) oxide

Cu(s) + O2(g) Cu2O(s)


Cu(s) + O2(g) Cu2O(s)

Topic: Writing Equations

Objective: How do we write equations when only given names?



3. Lead (II) nitrate + sodium phosphate lead (II) phosphate + sodium nitrate

Pb(NO3)2(aq) + Na3PO4(aq) Pb3(PO4)2(s) + NaNO3(aq)


Pb(NO3)2(aq) + Na3PO4(aq) Pb3(PO4)2(s) + NaNO3(aq)

Hey! With a little practice, this is easy and fun!



The mass on the reactants (left) side of the arrow and the mass on the

products (right) side of the arrow MUST equal each other as the Law of

Conservation of Mass states that mass may not be created or destroyed

in any chemical reaction or physical change.

Missing Mass examples:

1. Given that 35.0 g of N2(g) react with an unknown amount of H2(g) to

produce 42.5 g of NH3(g), how many grams of H2(g) were used?

i. 35.0 g of N2(g) + X g of H2(g) = 42.5 of NH3(g), therefore you first

solve for 42.5 g– 35.0 g = 7.5 g, so 7.5 g of H2(g) was used making

42.5 of NH3(g).

2. How many grams of aluminum are formed when 45.0 g of Al2O3(s)

are decomposed into Al(s) and 21.1 g of O2(g)?

i. 45.0 g of Al2O3 = 21.1 g of O2(g) + X g of Al(s), therefore you first

solve for 45.0 g – 21.1 g = 23.9 g, so 23.9 g of Al(s) is formed.

3. How many grams of FeS(s) must decompose to form 33.0 g of Fe(s)

and 19.0 g of S(s)?

i. 33.0 g of Fe (s) + 19.0 g of S(s) = X g of FeS(s), therefore you first

solve for X = 52.0 g, so 52.0 g of FeS(s) is decomposed.

Topic: Missing Mass in Equations

Objective: The amount of matter in reactants equals that in products!



Practice Regents Questions (ungraded):

1. During all chemical reactions, mass, energy, and charge are

a) Formed

b) Released

c) Absorbed

d) Conserved

2. Which of the following equations show conservation of mass?

a) H2 + O2 H2O

b) H2 + O2 2 H2O

c) 2 H2 + O2 2 H2O

d) 2 H2 + 2 O2 2 H2O

3. Which of the following equations show a conservation of mass?

a) H2O H2 + O2

b) PCl5 PCl3 + Cl2

c) Na + Cl2 NaCl

d) Al + Br2 AlBr3

4. Which of the following chemical equations is correctly balanced?

a) 2 KCl(s) 2 K(s) + Cl2(g)

b) H2(g) + O2(g) H2O(g)

c) N2(g) + H2(g) NH3(g)

d) 2 NaCl(s) Na(s) + Cl2(g)

Given the following unbalanced equation:

_1_ Fe2O3(s) + _3_ CO(g) _2_ Fe(s) + _3_ CO2(g)

5. When the equation is correctly balanced using the smallest whole-number

coefficients, what will be the coefficient of CO(g)?

a) 1 b) 2 c) 3 d) 4

Given the following unbalanced equation:

_1_ Mg(ClO3)2(s) _1_ MgCl2(s) + _3_ O2(g)

6. What is the coefficient of O2(g) when the equation is balanced correctly using the

smallest whole number coefficients?

a) 1 b) 2 c) 3 d) 4



Student name: _________Key__________ Class Period: _3, 5, & 10_

Please carefully remove this page from your packet to hand in.

Chemical Equations homework

What do each of the following mean in a chemical equation? 1 pt. ea.

(s) _solid_ (l) _liquid_ (g) __gas__ (aq) _aqueous_

Balance ALL the following equations/reactions by placing a coefficient in front of

the formulas. Do NOT change any subscript or element of each formula of a

compound. You don’t need to put (1) as a coefficient, but may do so if it is

helpful. 1 pt. ea.

Balance the reaction (Fill in the coefficients) 1 pt. ea.

_1_ C(s) + _2_ H2(g) _1_ CH4(g)

_2_ AgNO3(aq) + _1_ Cu(s) _2_ Ag(s) + _1_ Cu(NO3)2(aq)

_1_ N2(g) + _2_ O2(g) _1_ N2O4(g)

_4_ H2O2(l) _4_ H2O(l) + _2_ O2(g) simplify coefficients to _2_ _2_ + _1_

_1_ Ca(OH)2(aq) + _2_ HCl(aq) _1_ CaCl2(aq) + _2_ H2O(l)

Write the formulas for the following compounds. 1 pt. ea.

Compound name Compound formula Compound name Compound formula

Sodium oxide Na2O Lead (II) carbonate PbCO3

Potassium sulfate K2SO4 Lead (IV) carbonate Pb(CO3)2

Iron (II) chloride FeCl2 Zinc phosphate Zn3(PO4)2

Iron (III) chloride FeCl3 Zinc phosphide Zn3P2



Using the given names, write the formulas and balance the reactions. 1 pt. ea.

Iron + nitrogen gas iron (II) nitride

_3_ Fe(s) + _1_ N2(g) _1_ Fe3N2(s)

Lead + copper (II) nitrate lead (II) nitrate + copper

_1_ Pb(s) + _1_ Cu(NO3)2(aq) _1_ Pb(NO3)2(aq) + _1_ Cu(s)

Calcium + water calcium hydroxide + hydrogen gas

_1_ Ca(s) + _2_ H2O(l) _1_ Ca(OH)2(aq) + _1_ H2(g)

Potassium phosphate + iron (II) nitrate potassium nitrate + iron (II) phosphate

_2_ K3PO4(aq) + _3_ Fe(NO3)2(aq) _6_ KNO3(aq) + _1_ Fe3(PO4)2(s)

Find the Missing Mass for each of the problems below. 1 pt. ea.

If 15.0 g of nitrogen gas reacts with 3.2 g of hydrogen gas, how many grams of

ammonia gas will be formed?

_1_ N2(g) + _3_ H2(g) _2_ NH3(g)

15.0 g of N2(g) + 3.2 g of H2(g) = 18.2 g of N & H 18.2 g of NH3(g)

If 37.0 g of zinc(s) metal reacts with hydrochloric acid (HCl(aq)) to form 77.2 g of

zinc chloride(aq) and 1.1 g of hydrogen gas, how many grams of HCl were reacted?

_1_ Zn(s) + _2_ HCl(aq) _1_ ZnCl2(aq) + _1_ H2(g)

37.0 g of Zn(s) + X g of HCl = 77.2 g of ZnCl2(aq) + 1.1 g of H2(g)

(-37.0) X g of HCl = 77.2 + 1.1 - (37.0) = 41.3 g of HCl

How many grams of solid aluminum oxide must decompose to form 112.0 g of

solid aluminum metal and 99.6 g of oxygen gas?

_2_ Al2O3(s) _4_ Al(s) + _3_ O2(g)

X g of Al2O3(s) = 112.0 g of Al(s) + 99.6 g of O2(g)

211.6 g of Al2O3(s) = 112.0 + 99.6



Driving Force:

In nature, changes that require the least amount of energy will be the

changes that occur. Think of a rock; if you let go, it falls. Gravity is the

driving force behind the rock falling; no energy is required on the rock’s

part. The same premise is behind chemical reactions. Reactions will

occur in the direction that requires the least energy to occur.

Entropy as a Measure of Disorder:

Perhaps the best way to understand entropy as a driving force in nature

is to conduct a simple experiment with a new deck of cards. Open the

deck, and remove the jokers. The top card will be the ace of spades,

followed by the two, three, and four of spades, and so on. Now shuffle

the deck, and note that the deck becomes more disordered. The more

often the deck is shuffled, the more disordered it becomes. What makes

a deck of cards become more disordered when shuffled? The force

behind this increased disarray is entropy. The more disorder there is

the more entropy is in the system. If you picked up all the cards and put

them back into the box without sorting, they would be in disorder.

Sorting all the cards requires energy to be input. Reversing entropy

requires greater energy than it did to originally occur.

Topic: Driving Force and Entropy

Objective: What ‘drives’ chemical reactions to occur?



Determining Molecular Polarity:

Redox reactions:

Redox reactions are driven by the loss of electrons (oxidation) and the

gain of electrons (reduction).

In a redox reaction, one species (an element) gains electrons and the

other species loses electrons. Any ions of elements that are NOT

involved (no loss/no gain of e-) are called “spectator ions”.

Redox (Reduction-Oxidation): A reaction prompted by an exchange

in electrons between two elements, resulting in a change of the

oxidation numbers of both elements.

i. The more electronegative element (usually a nonmetal) gains the

electron(s), resulting in a decrease in oxidation number (becoming

more negative). This is reduction.

ii. The less electronegative element (usually a metal) loses the

electron(s), resulting in an increase in oxidation number (becoming

less negative). This is oxidation.

Watch YouTube Bozeman Science RedOx Reactions video (11:40)

https://www.youtube.com/watch?v=RX6rh-eeflM

Topic: Redox Reactions

Objective: How do we make the chemicals of everyday use?





Synthesis Redox Reactions:

*Note: When an element is shown as X0

, it means the charge is neutral.

Synthesis is the creation of something new from separate parts.

In a synthesis redox reaction, two elements (reactants) combine to

form a new compound (product).

This is the easiest way to make a new compound, and is often used in

industrial processes.

i. A + B AB

ii. A0 + B

0 A

+B

- (A

0 gives its electron to B

0; A

0 is oxidized, B

0 is

reduced)

Sample synthesis reaction (Rx):

i. 2 K + Br2 2 KBr Rx w/ charges: 2 K0 + (Br

0 + Br

0) 2 K

+ 2 Br

-

ii. K0 goes from 0 to +1, so K

0 is oxidized. K

0 gives electrons to Br

0.

Br0 goes from 0 to -1, so Br

0 is reduced.

Watch YouTube Swarthout Synthesis reaction video (6:49)

https://www.youtube.com/watch?v=MhlWTZwDHM8

Watch YouTube Swarthout Decomposition reaction video (8:13)

https://www.youtube.com/watch?v=1ocQhkHw_MM

Topic: Synthesis Reaction

Objective: How do we make a new compound with a redox reaction?







Decomposition Redox Reactions:

Decomposition is taking a compound and breaking it down to its

individual elements it was created from.

Decomposition redox reactions are rare in nature. This is due to the

fact that this requires more energy to break down compounds than

it took to form the compound. Water does not spontaneously break

down into hydrogen gas and oxygen; table salt does not

spontaneously break down into sodium metal and chlorine gas. Both

decompositions require a lot of energy.

Decomposition reactions are used to get pure highly reactive elements

(sodium, oxygen, chlorine, etc.) out of the naturally occurring

compounds where they are found.

Decomposition is often done using electrolytic decomposition, or

using electricity to break down a compound.

i. AB A + B

ii. A+B

- A

0 + B

0 (B

- gives its electron to A

+, so A

0 is reduced and

B0 is oxidized.)

Sample decomposition reaction (Rx):

i. 2 AgCl 2 Ag + Cl2 Rx w/ charges: 2 Ag+1

Cl-1

2 Ag0 + Cl2

0

ii. Ag+1

went from +1 to 0, so Ag+1

is reduced. Ag+1

gains electrons

from Cl-1

; Cl-1

goes from -1 to 0 and is oxidized.

Topic: Decomposition Reaction

Objective: How do we break down a compound into its elements?



Single Replacement Reaction:

Single replacement reactions have one single element and one type of

compound on either side of the reaction arrow.

o There are two possibilities of Single Replacement Reactions:

i. A metal plus a compound;

ii. A nonmetal plus a compound.

Topic: Single Replacement Reaction

Objective: How do we work with a single element and compound?

http://www.google.com/url?sa=i&rct=j&q=single+replacement+reaction+examples&source=images&cd=&cad=rja&uact=8&ved=0CAcQjRw&url=http://imgarcade.com/1/single-replacement-examples/&ei=4FbSVOyyI4-YyASZpoCICw&bvm=bv.85076809,d.aWw&psig=AFQjCNH0W1XPdwVFx-W_iTPWxDo8uExL1g&ust=1423157307549222



Single Replacement Reaction with a Metal:

1. Metal plus a compound single replacement.

i. The metal replaces the positive ion in the compound.

ii. This reaction may be used in batteries to create an electrical

current.

iii. This reaction may also be used to remove a less reactive element

from a compound.

iv. Adding a more reactive metal to the compound containing the less

reactive metal will “drive” the less reactive metal out, allowing

the more reactive metal to become part of the compound. The less

reactive metal will now be in its pure form.

v. This process is sometimes used in extraction of metal ore.

a. A + BC AC + B

b. A0 + B

+C

- B

+A

- + C

0 (C

- gives its electrons to A

0, so C

- is

oxidized and A0 is reduced.) B

+ does not have a change in

charge, so B+

is a spectator ion.

Sample Single Replacement Reaction with a Metal:

i. Reaction: Zn + Cu(NO3)2 Zn(NO3)2 + Cu

ii. Rx w/ charges: Zn0 + Cu

+2 (NO3)2

-1 Zn

+2(NO3)2

-1 + Cu

0

Topic: Single Replacement - Metal

Objective: How do we replace positive ions within a compound?



iii. Zn0 goes from 0 to +2, so it is oxidized. Zn

0 loses electrons to

Cu+2

, so Cu+2

goes from +2 to 0 and is reduced. NO3-1

does not

have a change in charge, so NO3-1

is a spectator ion.

Watch YouTube Swarthout Single Replacement reaction video (9:27)

https://www.youtube.com/watch?v=qhmTXdOKTBo





Single Replacement Reaction with a Nonmetal:

1. Nonmetal plus a compound single replacement.

i. The nonmetal replaces the negative ion in the compound.

ii. This is a rare type of reaction, and is a more costly method to

produce a pure nonmetal than using electrolytic decomposition.

a. A + BC BA + C

b. A0 + B

+C

- B

+A

- + C

0 (C

- gives its electrons to A

0, so C

- is

oxidized and A0 is reduced.) B

+ does not undergo a change in

charge, so B+

is a spectator ion.

Sample Replacement Reaction with a Nonmetal:

i. Reaction: F2 + ZnCl2 ZnF2 + Cl2

ii. Reaction w/ charges: F20 + Zn

+2Cl2

-1 Zn

+2F2

-1 + Cl2

0

iii. F0 goes from 0 to -1, so F

0 is reduced. F

0 gains electrons from the

Cl-1

, which goes from -1 to 0 and Cl-1

is oxidized. Zn+2

will not

have a change in charge, so Zn+2

is a spectator ion.

Topic: Single Replacement-Nonmetal

Objective: How do we replace negative ions within a compound?



Identifying the type of Reaction given:

1. 2 K + Zn(NO3)2 2 KNO3 + Zn

This reactions starts with an element and a compound and ends

with an element and a compound; it must be a single replacement

reaction.

2. 2 Fe + 3 Cl2 2 FeCl3

This reaction starts with elements and ends with a compound; it

must be a synthesis reaction.

3. 2 NaNO2 2 Na + N2 + 2 O2

This reaction starts with a compound and ends with elements; it

must be a decomposition reaction.

Topic: Identifying Reaction Type

Objective: How do identify the type of redox reaction?



Determining the charge of a reaction species:

i. Elements have NO charge; elements are neutral species.

ii. The charge for ions of elements (charged atoms) may be found

using the periodic table. If the metal (cation) has multiple charges

listed, use the nonmetal (anion) to determine the cation charge.

iii. Polyatomic ions have their charges listed in Reference Table E.

Examples:

1. Al2(SO4)3 ̶ Al is listed as forming only a +3

cation on the periodic

table, and the sulfate polyatomic ion is found as -2

on Table E as

SO4-2

. Each of the three polyatomic ion has a -2

charge, for a total

anion charge of -6, so the cations must equal a total +6 charge,

forcing the two Al atoms to form Al+3

ions:

Al2+3(SO4)3

-2

2. Fe3(PO4)2 ̶ Fe has two charges listed (+2

& +3

) on the periodic

table. The phosphate polyatomic ion is found as -3

on Reference

Table E as PO4-3

. As there are two phosphate polyatomic ions,

each with a -3

charge, the total negative charge is -6. Therefore the

total positive charge MUST be +6. As there are 3 Fe atoms, each

Fe atom needs to have a +2

charge to create the total +6 charge:

Fe3+2(PO4)2

-3

Topic: Determining Species Charge

Objective: How we determine the charges of the reaction species?



3. 4 Ag0 + O2

0 2 Ag2

+1O

-2 ̶ Note that the elements (Ag & O) are

by themselves and each have a charge of 0. O2 is a diatomic

molecule; as such it is neutral (charge of 0).

4. 6 K0 + Al2

+3(SO4)3

-2 3 K2

+1SO4

-2 + 2 Al

0 ̶ Note that the SUM of

the ion charges in each of the compounds totals ZERO.

5. 2 Fe+2

O-2

2 Fe0 + O2

0

6. 2 Na0 + Br2

0 2 Na

+1Br

-1

7. Ca0 + 2 H

+1Cl

-1 Ca

+2Cl2

-1 + H2

0

8. 2 Pb+2

O-2 2 Pb

0 + O2

0

9. 2 Al0 + 3 Ag2

+1CO3

-2 Al2

+3(CO3)3

-2 + 6 Ag

0



Determining Species Type in a Reaction:

i. The species in a reaction that is OXIDIZED lost electrons, and the

oxidation number has become more POSITIVE.

ii. The species in a reaction that is REDUCED gained electrons, and

the oxidation number has become more NEGATIVE.

iii. The species that is the SPECTATOR ION neither gains nor loses

electrons, and its charge remains the SAME.

Examples:

OX = Oxidized species RD = Reduced species SI = Spectator ion

1. 4 Ag0 + O2

0 2 Ag2

+1O

-2

OX (charge ↑ +): Ag0 RD (charge ↑ -): O2

0 SI: none

2. 6 K0 + Al2

+3(SO4)3

-2 3 K2

+1SO4

-2 + 2 Al

0

OX (charge ↑ +): K0 RD (charge ↑ -): Al

+3 SI: SO4

-2

Topic: Determining Species Type

Objective: How do we find the oxidized, reduced, or spectator ion?



3. 2 Fe+2

O-2

2 Fe0 + O2

0

OX (charge ↑ +): O-2

RD (charge ↑ -): Fe+2

SI: none

4. 2 Na0 + Br2

0 2 Na

+1Br

-1

OX (charge ↑ +): Na0 RD (charge ↑ -): Br2

0 SI: none

5. Ca0 + 2 H

+1Cl

-1 Ca

+2Cl2

-1 + H2

0

OX (charge ↑ +): Ca0 RD (charge ↑ -): H

+1 SI: Cl

-1

6. 2 Pb+2

O-2

2 Pb0 + O2

0

OX (charge ↑ +): O-2

RD (charge ↑ -): Pb+2

SI: none

7. 2 Al0 + 3 Ag2

+1CO3

-2 Al2

+3(CO3)3

-2 + 6 Ag

0

OX (charge ↑ +): Al0 RD (charge ↑ -): Ag

+1 SI: CO3

-2



Determining the missing species in a reaction:

If the missing species is an ELEMENT, write its symbol. If the missing

species is diatomic (Br2, I2, N2, Cl2, H2, O2, F2), write it as a diatomic

molecule. This will balance the equation.

If the missing species is a COMPOUND, write the negative ion first and

the positive ion second. Write the formula in such a way that you have

the equal numbers of atoms of each element on both sides of the

balanced reaction.

Examples:

1. 4 Ag + O2 2 _ Ag2O _ (missing species is Ag2O)

i. There are now 4 Ag and 2 O on each side of the reaction arrow.

2. 6 K + Al2(SO4)3 3 _ K2SO4_ + 2 Al (missing species is K2SO4)

i. There are 6 K’s and 3 SO4’s missing, so with the coefficient of 3,

the new compound contains 2 K’s (3 x 2) and 1 SO4 (3 x 1).

ii. There are two Al’s on both sides…OK!

Topic: Finding a Missing Species

Objective: How do we find which part of a reaction is missing?



3. 2 FeO 2 Fe + ___ O2___ (missing species is O2)

i. O2 is diatomic, so there are now 2 O’s on each side.

ii. There are two Fe’s on both sides…OK!

4. 2 Al + 3 _ Ag2CO3_ Al2(CO3)3 + 6 Ag (Ag2CO3 is missing)

i. There are 6 Ag’s and 3 CO3’s missing. Since the coefficient is 3,

the compound contains 2 Ag’s (3 x 2) and 1 CO3 (3 x 1).

ii. There are two Al’s on both sides…OK!

5. 2 Na + Br2 2 _NaBr_ (missing species is NaBr)

6. Ca + 2 __HCl__ CaCl2 + H2 (missing species is HCl)

7. 2 PbO 2 ___Pb___ + O2 (missing species is Pb)



Regents Practice Problems (ungraded):

Given the balanced equation:

2 KClO3 2 KCl + 3 O2

1. Which type of reaction is represented by the above equation?

a) Synthesis

b) Decomposition

c) Single Replacement

d) Nuclear Transmutation

Given the balanced equation representing the reaction:

4 Al(s) + 3 O2(g) 2 Al2O3(s)

2. Which type of chemical reaction is represented by the above equation?

a) Synthesis

b) Decomposition



3. Which of the balanced equations below represent a chemical change?

a) H2O(l) H2O(s) + energy

b) H2O(g) H2O(l) + energy

c) H2O(l) + energy H2O(g)

d) 2 H2O(l) + energy 2 H2(g) + O2(g)

4. Which of the following lists includes three types of chemical reactions?

a) Condensation, solidification, and synthesis

b) Decomposition, solidification, and sublimation

c) Decomposition, single replacement, and synthesis

d) Condensation, single replacement, and sublimation

Given the balanced reaction:

Mg(s) + 2 AgNO3(aq) Mg(NO3)2(aq) + 2 Ag(s)

5. Which type of reaction is represented above?

a) Synthesis

b) Decomposition





Student name: _________Key__________ Class Period: _3, 5, & 10_


Oxidation and reduction Reactions homework

*Note: Make sure to include ALL charges when writing a species!

Write Al+3

, not Al Write Na0, not Na Write Cl

-1, not Cl

Multiple choice questions: circle the correct answer choice. 1 pt. ea.

1. In the reaction 2 Na + 2 HNO3 2 NaNO3 + H2, which species is the spectator

ion?

a) Na0 b) H

+1 c) NO3

-1 d) Na

+1

2. In the reaction Zn + 2 HNO3 ________ + H2, the missing species is:

a) Zn(NO3)2

b) (NO3)2Zn

c) ZnNO3

d) NO3Zn

3. In the reaction 2 ________ 2 Pb + O2, the missing species is:

a) PbO b) OPb c) PbO2 d) O2Pb

Identify the type of reaction (synth, decomp, or sngl rplcmnt) below. 1 pt. ea.

Given Reaction Reaction type How can you tell the type?

KBr + Na NaBr + K sngle rplcmnt Cmpd + Elmt Cmpd + Elmt

2 LiNO3 + Ca Ca(NO3)2 + 2 Li sngl replcmnt Cmpd + Elmt Cmpd + Elmt

2 Fe + 3 Cl2 2 FeCl3 synthesis Elmt + Elmt Cmpd

2 Li2O 4 Li + O2 decomp Cmpd Elmt + Elmt

Write the charges of each of the species in the reactions below, then identify which

species are oxidized, reduced, or are spectator ions (if any are present). 1 pt. ea.

Ag2S 2 Ag + S 2 Ag+1

S-2

2 Ag0 + S

0

Oxidized: ___S-2

___ Reduced: ___Ag+1

___ Spectator ion: __none__



KBr + Na NaBr + K K+1

Br-1

+ Na0 Na

+1 Br

-1 + K

0

Oxidized: ___Na0___ Reduced: ___K

+1___ Spectator ion: ___Br

-1___

2 LiNO3 + Ca Ca(NO3)2 + 2 Li 2 {Li

+1 NO3

-1} + Ca

0 Ca

+2 (NO3)2

-1 + 2 Li

0

Oxidized: ____Ca0____ Reduced: ___Li

+1___ Spectator ion: ___NO3

-1___

2 Fe + 3 Cl2 2 FeCl3 2 Fe

0 + 3 Cl2

0 2 {Fe

+3 Cl3

-1}

Oxidized: ____Fe0____ Reduced: ___Cl2

0___ Spectator ion: ___none___

Complete the following reactions by writing the appropriate formula and balancing

the equations. 1 pt. ea.

2 Na + Br2 2 __NaBr__

3 Zn + 2 Fe(NO3)3 3 __Zn(NO3)2__ + 2 Fe

2 BaO 2 Ba + __O2__

2 Sc + 3 CuSO4 Sc2(SO4)3 + 3 ____Cu____

2 NO2 ____N2____ + 2 O2

2 ____Li____ + S Li2S

Cu + 2 ___KNO2___ Cu(NO2)2 + 2 K

Mg + ____Cl2____ MgCl2

2 ___Li3N___ 6 Li + N2

2 Na + ___CaCO3___ Na2CO3 + Ca



Aqueous Ionic Compounds:

1. Soluble Ionic Compounds:

i. Ionic compounds generally ionize in water. When a soluble ionic

compound is placed into water, the interaction between the strong

polar bonds of the water molecule and the weaker bonds of the

ionic compound rips the ionic compound into its component ions.

ii. The more electronegative (δ-) oxygen end of the water molecule

attracts and attaches to the more positive ion. The less

electronegative (δ+) hydrogen end of the water molecule attracts and

attaches to the more negative ion. The liquid water molecules,

continually moving, tear the crystal ions apart and keep the ions

apart in solution.

2. Insoluble Ionic Compounds:

i. Insoluble compounds have ionic structures that remain together

when in water. This is because the attractions between the ions are

too strong for water molecules to tear them apart. Insoluble ions

attach to other insoluble ions, and the particles form a cloudy

suspension. The suspension will eventually form solid crystals,

called a precipitate, which settles due to gravity.

Topic: Solubility & Ionic Compounds

Objective: How do ionic substances interact in aqueous solutions?



Which Ionic Compounds are Soluble (or Insoluble) in water?

To determine if an ionic compound is soluble or insoluble in water,

use Reference Table F, Solubility Guidelines for Aqueous Solutions.

Note that there are two columns, Ions That Form Soluble

Compounds, and Ions That Form Insoluble Compounds in

Reference Table F. Note that to the right of each column is the

Exceptions column for either Soluble or Insoluble ions. The

Exceptions column gives examples when the general Soluble or

Insoluble guideline is NOT applicable for that type of ion. Be careful

when reading Table F; it can cause headaches!

Topic: Determining Solubility

Objective: How do we find which ionic compounds are soluble?



Use Reference Table F, Solubility Guidelines for Aqueous Solutions, to

determine the reason why a compound is soluble or insoluble in water.

Ionic

Compound

Soluble or

Insoluble? Reason why is Soluble or Insoluble.

Li2CO3 Soluble Li

+1 is a Group 1 ion, and ALL Group 1 ions are soluble

(blank exceptions)

Pb(NO3)2 Soluble NO3-1

(nitrate) is given as soluble (blank exceptions)

ZnCl2 Soluble Cl

-1 is a halide, given as soluble, with few exceptions

(Zn+2

is NOT an exception)

BaSO4 Insoluble SO4

-2 (sulfate) is given as soluble, with few exceptions

(Ba+2

IS an exception)

Ca(HCO3)2 Soluble HCO3

-1 (hydrogen carbonate) is given as soluble (blank

exceptions)

CuCO3 Insoluble

CO3-2

(carbonate) is given as insoluble, with few

exceptions (Cu+2

is NOT an exception, as Cu+2

is NOT a

Group 1 ion)

Pb(CrO4)2 Insoluble

CrO4-2

(chromate) is given as being insoluble, with few

exceptions (Pb+2

is NOT and exception, as Pb+2

is NOT a

Group 1 ion)

Na3PO4 Soluble PO4

-3 (phosphate) is insoluble EXCEPT with Group 1 or

NH4+1

(Na+1

is a Group 1 ion, so Na+1

IS an exception)

NH4OH Soluble NH4

+1 (ammonium) is given as always soluble (blank

exceptions)

Mg(OH)2 Insoluble

OH-1

(hydroxide) is given as insoluble, with few

exceptions (Mg+2

is NOT an exception, as Mg+2

is NOT a

Group 1 ion)

Topic: Using Table F for Solubility

Objective: How do we interpret Reference Table F?



Double Replacement Reactions:

In a double replacement reaction the positive ion of one ionic

compound swaps places with the negative ion from the other ionic

compound. The double replacement reaction is carried out within

aqueous solutions of each reactant ionic compound so that the ions

may more easily mix and react.

General Double Replacement Formula:

A+B-(aq) + C+D-

(aq) A+D-(s) + C+B-

(aq)

Watch YouTube Swarthout Double Replacement reaction video (10:49)

https://www.youtube.com/watch?v=7hVKb4ROjZw

+

Topic: Double Replacement Reaction

Objective: What happens in a double replacement reaction?

+





Aqueous solution AB (composed of separated A+ and B

- ions) is mixed

into aqueous solution CD (composed of separated C+ and D

- ions). As

they are mixed, the A+ ions of the first solution seek out and tightly bond

with the D- ions from the second solution. These A

+ and D

- ions bond so

tightly that that the polar water molecules cannot strip them apart. The

resulting ionic compound AD is formed of small crystals that turn the

combined solution cloudy. As the AD crystals clump together and gain

mass, they will settle out due to gravity. The settled solid AD is the

precipitate of the reaction.

Examples of Double Replacement Reactions:

1. K2CrO4(aq) + Ba(NO3)2(aq) 2 KNO3(aq) + BaCrO4(s)

i. Look for the products (K+ and NO3

-) and (Ba

+2 and CrO4

-2) in

Reference Table F. K+1

is a Group 1 ion; K+ is always soluble.

NO3- is given as always soluble as well. The KNO3 will remain as

aqueous (in solution with water). The CrO4-2

is listed as insoluble,

and Ba+2

is NOT one of the exceptions for CrO4-2

. Therefore the

BaCrO4 is the solid precipitate that forms from this reaction.

2. Na2CO3(aq) + CaCl2(aq) 2 NaCl(aq) + CaCO3(s)

i. Look for the products (Na+ and Cl

-) and (Ca

+2 and CO3

-2) in Table F.

Na+1

is another Group 1 ion, and always soluble. Cl-1

is usually

soluble, and Na+1

is not an exception here. The NaCl will remain as

aqueous. The CO3-2

is given as insoluble, and Ca+2

is NOT an

exception for CO3-2

. Therefore the CaCO3 is the solid precipitate

that forms from this reaction.



Completing a Double Replacement Reaction:

How may we complete a Double Replacement Reaction that contains

missing compounds? The same way we did earlier with single redox

reactions.

i. For the missing compound, write the negative ion first and the positive

ion second. Write the formula in such a way that you have the equal

numbers of atoms of each element on both sides of the balanced

reaction.

ii. Once you have the balanced reaction, use Reference Table F to

identify the precipitate in the reaction.

*NOTE: Not all double replacement reactions form precipitates!

Examples of Completing a Double Replacement Reaction:

I. KCl(aq) + AgNO3(aq) KNO3(_____) + _________(_____)

a) The missing compound will contain Ag+1

and Cl-1

. Since no

coefficient was given, the formula will be AgCl. Using

Reference Table F, KNO3 is soluble (will be aqueous), and the

AgCl(s) will be the insoluble precipitate.

KCl(aq) + AgNO3(aq) KNO3(__aq__) + __AgCl__(__s__)

Topic: Double Replacement Reaction

Objective: How can double replacement reactions be completed?



II. 2 Na3PO4(aq) + 3 Mg(NO3)2 6 NaNO3(______) + ___________(_____)

a) The missing compound contains Mg and PO4. The reactants

contain 2 PO4’s and 3 Mg’s. There is no coefficient given for the

missing product, so the formula is Mg3(PO4)2. Using Table F,

NaNO3 is soluble and will remain in aqueous solution, and

Mg3(PO4)2(s) will be the insoluble precipitate.

2 Na3PO4(aq) + 3 Mg(NO3)2 6 NaNO3(__aq__) + _Mg3(PO4)2_(__s__)

The Double Replacement reactions shown in examples I and II both

formed solid precipitates. While most double replacement reactions we

will work with do form solid precipitates, two other double replacement

reactions do NOT form precipitates. They are neutralization and gas-

forming reactions.

If solution AB is an acid (consisting of AH+ and BNnM

- aqueous ions)

and solution CD is a base (consisting of CM+ and DOH

- ions). When a

double displacement reaction occurs, the cations and anions switch

partners, resulting in the formation of water and a new ionic

compound (or salt), which is usually soluble.

Example of a neutralization reaction:

a) Sulfuric acid(aq) + lithium hydroxide(aq) water(l) + lithium sulfate(aq)

b) Equation: H2SO4(aq) + 2 LiOH(aq) Li2SO4(aq) + 2 H2O(l)



Spectator Ion Determination:

In a Double Replacement Reaction the ions that remain dissolved in

the solution (that remain aqueous) are called Spectator Ions.

We may identify the Spectator Ions by determining which ions

remain in aqueous solution on BOTH sides of the equation.

Mg(NO3)2(aq) + Na2CO3(aq) MgCO3(s) + 2 NaNO3(aq)

Compare the reactants (right side) with the products (left side):

i. Mg+2

is found in the product precipitate; it underwent a replacement.

ii. NO3-1

is found in aqueous compounds on BOTH sides; it was

unchanged.

iii. Na+1

is found in aqueous compounds on BOTH sides; it was

unchanged.

iv. CO3-2

is found in the product precipitate; it underwent a

replacement.

v. As the NO3-1

and Na+1

are the only ions that started in aqueous

solution and finished in aqueous solutions, they are the spectator

(unchanged) ions in this precipitate reaction.

Topic: Identifying Spectator Ions

Objective: How do we find the aqueous ions that remain unchanged?



Student name: __________Key__________ Class Period: _3, 5, & 10_


Double Replacement Homework

Using Reference Table F, determine if the following compounds are soluble or

insoluble and give your reason(s). 1 pt. ea.

Compound

Formula

Soluble or

Insoluble Reasons for your choice

Na2CO3 S Sodium is a Group 1 exception for CO3-2

CaS I Sulfide is insoluble with all but Grp 1 & NH4+1

PbBr2 I Halides (Br) are insoluble with lead (II)

Al(OH)3 I Hydroxide is insoluble with all but Grp 1, some Grp 2, & NH4+1

(NH4)2CrO4 S Ammonium is listed as exception to insoluble CrO4-2

CIRCLE the precipitate (insoluble product) in each double replacement reaction.

Reaction (Circle each precipitate compound in the product) 1 pt. ea.

3 Ca(NO3)2(aq) + 2 K3PO4(aq) 6 KNO3(aq) + Ca3(PO4)2(s)

AgNO3(aq) + NaOH(aq) NaNO3(aq) + AgOH(s)

Na2SO4(aq) + Ba(ClO3)2(aq) NaClO3(aq) + BaSO4(s)

(NH4)2CO3(aq) + Ca(C2H3O2)2(aq) CaCO3(s) + 2 NH4C2H3O2(aq)

2 AgNO3(aq) + Li2CO4(aq) 2 LiNO3(aq) + Ag2CrO4(s)

Write the formula for the missing product (give as (s) or (aq) as well) in these double

replacement reactions. 2 pts ea.

i. CaCl2(aq) + Na2CO3(aq) CaCO3(s) + 2 _____NaCl(aq)_____

ii. (NH4)2CO3(aq) + Cu(NO3)2(aq) CuCO3(s) + 2 ____NH4NO3(aq)____

iii. 2 Li3PO4(aq) + 3 Fe(NO3)2(aq) 6 LiNO3(aq) + ___Fe3(PO4)2(s) ___

iv. Pb(ClO3)2(aq) + 2 KBr(aq) PbBr2(s) + 2 ____KClO3(aq)___



For each of the multiple choice questions below circle the correct answer. 1 pt. ea.

1. Which of the following equations represent a double replacement reaction?

a) 2 H2(g) + O2(g) 2 H2O(l)

b) CaCO3(s) CaO(s) + CO2(g)

c) KOH(aq) + HCl(aq) KCl(aq) + H2O(l) Acid-Base Neutralization Rx

d) Cu(s) + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq)

Given the balanced equation:

AgNO3(aq) + NaCl(aq) NaNO3(aq) + AgCl(s)

2. The reaction above may best be classified as

a) Synthesis

b) Decomposition


d) Double Replacement

3. Which of the following equations represent a double replacement reaction?

a) CaCO3(s) CaO(aq) + CO2(g)

b) LiOH(aq) +HCl(aq) LiCl(aq) + H2O(l) Acid-Base Neutralization Rx

c) CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)

d) 2 Na(s) + 2 H2O(l) 2 NaOH(aq) + H2(g)

Given the balanced reaction:

Ba(NO3)2(aq) + Na2SO4(aq) 2 NaNO3(aq) + BaSO4(s)

4. Which of the following is the name for the precipitate formed?

a) Nitrate oxide

b) Sodium oxide

c) Barium nitrate

d) Barium sulfate



A chemical reaction is basically a recipe used to make something new

from “ingredients”. Just like any other recipe, we can make a smaller or

larger amount by changing the amount of the reactants used.

For chemical reactions the coefficients in front of each species tells you

what the proportional number of moles for each reactant is needed to

make the proportional number of moles of product. The reaction

below is the Haber Process to manufacture ammonia:

N2(g) + 3 H2(g) 2 NH3(g)

The reaction above states that when nitrogen and hydrogen are reacted in

a 1:3 mole ratio, you will produce 2 moles of ammonia. The same 1:3

ratio applies to a small batch of ammonia, or a larger batch. Therefore,

if asked to produce a total of 10 moles of ammonia, you can easily scale

up the “recipe” to make what is asked.

When you divide the “asked” amount (10 moles) by the coefficient for

the reaction (2 moles), you have a factor of 5. This factor of 5 is then

multiplied by the reactants (5 x 1 mole of N2(g)) and (5 x 3 moles of

H2(g)) which will result in the production of 10 moles of NH3(g).

(5 x 1) moles of N2 needed (5 x 3) moles of H2 needed Asked for 10 moles of NH3 (10/2 moles) = 5x

1 N2(g) + 3 H2(g) 2 NH3(g)

Topic: Stoichiometry of Equations

Objective: How do we make the amount of product needed?



Question asked: How many moles of product ‘X’ are formed when ‘n’ moles of ‘Y’ react?

How may we use this simple mole ratio relationship to determine what

to use in a chemical reaction? There is a very simple equation to use:

Moles of given x 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡𝑎𝑟𝑔𝑒𝑡

𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑔𝑖𝑣𝑒𝑛 = Moles of target

Note that for the equation above and for each of the following examples,

the “given” mole ratio is written in strikethrough font, indicating that

this unit cancels out to leave the target.

Stoichiometry Equation examples:

For the reaction N2(g) + 3 H2(g) 2 NH3(g), how many moles of NH3(g) will be formed

when 6.0 moles of N2(g) are completely reacted with excess (more than needed) H2(g)?



6.0 moles of N2(g) x 2 moles of NH3(g)

= 12. moles of NH3(g) 1 mole of N2(g)

6.0 moles of N2(g) reacted with excess H2(g) will form 12. moles of NH3(g)

For the reaction N2(g) + 3 H2(g) 2 NH3(g), how many moles of H2(g) will be needed to

completely react with excess (more than needed) N2(g) to form 1000. moles of NH3(g)?



1000. moles of NH3(g) x 3 moles of H2(g)

= 1500. moles of H2(g) 2 moles of NH3(g)

1500. moles of H2(g) reacted with excess N2(g) will form 1000. moles of NH3(g)

Topic: Stoichiometry of Equations

Objective: How do we make the amount of product needed?



For the reaction 2 Na(s) + 2 H2O(l) 2 NaOH(aq) + H2(g), how many moles of Na(s) are

needed to form 4.0 moles of H2(g)?



4.0 moles of H2(g) x 2 moles of Na(s)

= 8.0 moles of Na(s) 1 mole of H2(s)

8.0 moles of Na(s) are reacted with excess H2O(l) and NaOH(aq) to form 4.0 moles of H2(g)

For the reaction 4 Al(s) + 3 O2(g) 2 Al2O3(s), how many moles of Al2O3(s) will form if 6.0

moles of Al(s) are reacted in excess O2(g)?



6.0 moles of Al(s) x 2 moles of Al2O3(s)

= 3.0 moles of Al2O3(s) 4 moles of Al(s)

3.0 moles of Al2O3(s) will be formed if 6.0 moles of Al(s) react with excess O2(g)

For the reaction 2 NaCl(aq) + Pb(NO3)2(aq) 2 NaNO3(aq) + PbCl2(s), how may moles of

PbCl2(s) precipitate will form when 5.0 moles of Pb(NO3)2(aq) are reacted in excess with

NaCl(aq)?



5.0 moles of Pb(NO3)2(aq) x 1 mole of PbCl2(s)

= 5.0 moles of PbCl2(s) 1 mole of Pb(NO3)2(aq)

5.0 moles of PbCl2(s) will form when 5.0 moles of PbCl2(s) react in excess NaCl(aq)

Watch YouTube Bozeman Science Stoichiometry video (9:45)

https://www.youtube.com/watch?v=LQq203gyftA





Practice Regents Questions (ungraded):

Given the balanced equation for the reaction:

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

1. What is the total number of moles of O2(g) required for the complete combustion

(reaction) of 1.5 moles of C3H8(g)?

a) .30 moles of O2(g)

b) 1.5 moles of O2(g)

c) 4.5 moles of O2(g)

d) 7.5 moles of O2(g)

Given the balanced equation for the reaction:

2 CO(g) + O2(g) 2 CO2(g)

2. What is the mole ratio of CO(g) to CO2(g) in this reaction?

a) 1:1

b) 1:2

c) 2:1

d) 3:2

Given the balanced reaction between the gases propane and oxygen:

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

3. Which of the following shows the correct fractional ratio of oxygen to propane?

a) 5 grams of O2

c) 10 grams of O2

1 gram of C3H8 11 grams of C3H8

b) 5 moles of O2

d) 10 moles of O2

1 mole of C3H8 11 moles of C3H8

Given the reaction:

C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l)

4. How many moles of C6H12O6(s) are needed to produce 24 moles of carbon

dioxide gas?

a) 1.0 moles

b) 4.0 moles

c) 12. moles

d) 24. moles



Student name: _________________________ Class Period: _______


Stoichiometry of Equations homework

For ALL mole conversion problems given here, show ALL of your work,

including showing which units cancel by lining them out.

1. Given the reaction of N2(g) + H2(g) NH3(g), answer the following. 1 pt. ea.

i. Balance the reaction: __1__ N2(g) + __3__ H2(g) __2__ NH3(g)

ii. How many moles of N2(g) are needed to form 5.0 moles of NH3(g) reacted in excess H2(g)?

5.0 moles of NH3(g) x 1 mole of N2(g)

= 2.5 moles of N2(g) needed 2 moles of NH3(g)

iii. How many moles of N2(g) are needed to completely react with 10.0 moles of H2(g)?

10.0 moles of H2(g) x 1 mole of N2(g)

= 3.33 moles of N2(g) needed 3 moles of H2(g)

iv. How many moles of NH3(g) should form from 6.0 moles of H2 reacted in excess N2(g)?

6.0 moles of H2(g) x 2 moles of NH3(g)

= 4.0 moles of NH3(g) needed 3 moles of H2(g)

2. Given the reaction of Zn(s) + HBr(aq) ZnBr2(aq) + H2(g), answer the following. 1 pt. ea.

i. Balance the reaction: __1__ Zn(s) + __2__ HBr(aq) __1__ ZnBr2(aq) + __1__ H2(g)

ii. How many moles of Zn(s) are needed to form 8.0 moles of ZnBr2(aq) in excess HBr(aq)?

8.0 moles of ZnBr2(aq) x 1 mole of Zn(s)

= 8.0 moles of Zn(s) needed 1 mole of ZnBr2(aq)

iii. How many moles of HBr(aq) are needed to form 4.0 moles of H2(g)?

4.0 moles of H2(g) x 2 moles of HBr(aq)

= 8.0 moles of HBr(aq) needed 1 mole of H2(g)

iv. How many moles of ZnBr2(aq) will form if 5.0 moles of HBr(aq) are reacted in excess Zn(s)?

5.0 moles of HBr(aq) x 1 moles of ZnBr2(aq)

= 2.5 moles of ZnBr2(aq) needed 2 moles of HBr(aq)



3. Given the reaction of Ca(s) + N2(g) Ca3N2(s), answer the following. 1 pt. ea.

i. Balance the reaction: __3__ Ca(s) + __1__ N2(g) __1__ Ca3N2(s)

ii. How many moles of Ca(s) are required to form 4.00 moles of Ca3N2(s) in excess N2(g)?

4.00 moles of Ca3N2(s) x 3 moles of Ca(s)

= 12.0 moles of Ca(s) needed 1 mole of Ca3N2(s)

iii. How many moles of Ca3N2(s) will form if 2.50 moles of N2(g) react with excess Ca(s)?

2.50 moles of N2(g) x 1 mole of Ca3N2(s)

= 2.50 moles of Ca3N2(s) needed 1 mole of N2(g)

iv. How many moles of Ca(s) are needed to completely react with 5.00 moles of N2(g)?

5.00 moles of N2(g) x 3 moles of Ca(s)

= 15.0 moles of Ca(s) needed 1 mole of N2(g)

4. Given the reaction of Al(s) + MgCl2(aq) AlCl3(aq) + Mg(s), answer the following. 1 pt. ea.

i. Balance the equation: __2__ Al(s) + __3__ MgCl2(aq) __2__ AlCl3(aq) + __3__ Mg(s)

ii. How many moles of Mg(s) will form if 35.0 moles of Al(s) react in excess MgCl2(aq)?

35.0 moles of Al(s) x 3 moles of Mg(s)

= 52.5 moles of Mg(s) needed 2 moles of Al(s)

iii. How many moles of MgCl2(aq) are required to completely react with 15.0 moles of Al(s)?

15.0 moles of Al(s) x 3 moles of MgCl2(aq)

= 22.5 moles of MgCl2(aq) needed 2 moles of Al(s)

iv. How many moles of AlCl3(aq) will be formed if 13.0 moles of Mg(s) are formed?

13.0 moles of Mg(s) x 2 moles of AlCl3(aq)

= 8.66 moles of AlCl3(aq) needed 3 moles of Mg(s)



Notes page:

unit 8: reactions-key regents chemistry ’14 mr. · pdf fileunit 8: reactions-key regents...

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