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Chapter 9

Reaction Stoichiometry

What is Reaction Stoichiometry???

Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet

“In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.”

Reaction Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

Review: Chemical EquationsChemical change involves a reorganization of the atoms in one or more substances.

C2H5OH + 3O2 → 2CO2 + 3H2Oreactants products

1 molecule of ethanol reacts with 3 molecules of oxygen to produce 2 molecules of carbon dioxide and 3 molecules of water

When the equation is balanced it has quantitative significance:

C2H5OH + 3O2 → 2CO2 + 3H2Oreactants products

1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

In Stoichiometry the equation is read as:

Mole Relations

What are mole ratios and how are they used in stoichiometric calculations?

Mole ratios are conversion factors.

Mole ratios are formed using coefficients.

Mole ratios will help predict amounts of R’s and P’s from a balanced equation.

Form mole ratios based on the following equation.

EQUATIONS MUST BE BALANCED!!!!!!!!!!!

Ex: N2 (g) + 3H2 (g) 2NH3 (g) Write all possible mole ratios for the above equation.

Mole RatiosN2 (g) + 3H2 (g) 2NH3 (g)

Write all possible mole ratios for the following equations.

2HgO (s) 2Hg (l) + O2 (g)

4NH3 (g) + 6NO (g) 5N2 (g) + 6H2O (l)

2Al (s) + 3H2SO4(aq) Al2(SO4)3 + 3H2 (g)

Mole Ratios2HgO (s) 2Hg (l) + O2 (g)

Mole Ratios4NH3 (g) + 6NO (g) 5N2 (g) + 6H2O (l)

Mole Ratios2Al (s) + 3H2SO4(aq) Al2(SO4)3 + 3H2 (g)

Mole ratios can be interpreted as:

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

Are molecules conserved??NO

Particles Are particles conserved?? (Avogadro’s #)NO

Volume (ch 10-skip) Is volume conserved??NO

Moles Are moles conserved?? NO

Mass Is mass conserved??YES – Law of conservation of mass

Atoms Are atoms conserved??YES –EQUATIONS MUST BE BALANCED!!

Reaction Stoichiometry deals with mass (quantities) relationships

Quantities can be measured in either moles or grams.

Grams are measurable in lab settings.

Take notes on all sample problems before reading the “story”.

Sample Calculations: 4 General Types Type #1: mole mol

Strategy: Write given amounts above equations and molar masses below

Ammonia, NH3 is widely used a s a fertilizer and in many household cleaners. How many moles of ammonia are produced when 6 mol of hydrogen gas react with an excess of nitrogen gas? 6mol excess ? mol

Equation: 3H2 + N2 2NH3

?mol NH3 = 6 mol H2 2mol NH3 3mol H2

= 4 mol NH3

Type # 2: mol gram

When magnesium burns in air, it combines with oxygen to form magnesium oxide according to the following equation:

2.00 moles ?g 2 Mg (s) + O2 (g) 2 MgO (s)

40.31 g

What mass (g) of magnesium oxide is produced from 2.00 mol of magnesium?

? g MgO = 2.00 mole Mg 2 mol MgO 2 mol Mg

40.31 g MgO 1 mol MgO

= 80.6 g MgO

Type #3: gram mol

Oxygen was discovered by Joseph Priestley in 1774 when he heated mercury (II) oxide to decompose it to form its constituent elements. How many moles of mercury (II) oxide, HgO, are needed to produce 125 g of oxygen, O2?

?mol 125 g 2 HgO 2Hg + O2

32.00 g = molar mass

? mol HgO = 125 g O2 1mol O2 32.00 g O2

2 mol HgO1 mol O2

= 7.81 mol HgO

Type #4: gram gram

Laughing gas (nitrous oxide, N2O) is sometimes used as an anesthetic in dentistry. It is produced when ammonium nitrate is decomposed according to the following reaction:

? g 33.0 gNH4NO3 (s) N2O (g) + 2H2O (l)

80.06 g = molar mass 18.02 g = molar mass

How many grams of NH4 NO3 are required to produce 33.0 g of H2O?

? g NH4NO3= 33.0 g H2O 1 mol H2O

18.02 g H2O1 mol NH4 NO32 mol H2O

80.06 g NH4NO3 1 mol NH4NO3

= 73.3 gNH4NO3

Moles of Given   Moles of Wanted    

Molar MASS

Mole Ratio

Type 3Mass of Given              Moles of Given  Moles of Wanted  Mole Ratio 

Moles of Given Mole Ratio Molar MASS Mass of Wanted

Molar MASSMoles of Given Mole Ratio Mass of WantedMass of Given Molar MASS

Type 1

Type 2

Type 4

Moles of Wanted

Moles of Wanted

Story

Once upon a time, YOU lived in a town by the name of GIVEN.You had friends in a town nearby called

ASKED FOR.One day you decided to visit your friends in

ASKED FOR,but

The only way to get from GIVEN to ASKED FOR

was to cross the “OLD MOLE BRIDGE”The reason they call it the“Old Mole Bridge”…..

Story continued…is because only MOLES are allowed to cross it.

So, the only way to get to ASKED FOR

is to change yourself into a MOLE

And cross the bridge.Of course, you don’t want to be a MOLE for any longer

than you have to,So after crossing the bridge, you change yourself back, and now

you’ve made it to ASKED FOR.

Limiting Reactant

Limiting Reactant – the reactant that is consumed (runs out) first.Limiting Reactant - limits the amounts of products formed.

Limiting Reagents - Combustion

Excess ReactantExcess Reactant – the reactant that does not run out. Some remains at the end of the reaction.

Steps for determining limiting and excess reactants.

Write a balanced chemical equation.

Identify both given quantities.

Based on both given quantities, solve for one of the products twice.

Cross out the larger amount of the product produced. It is not formed

This indicates the excess reactant. Label the excess.

The smaller amount of the product indicates the limiting reactant. Label the limiting.

The smaller amount of the product is the theoretical yield. Label the theoretical yield.

Always use the limiting reactant to solve for other steps including excess reactant remaining.

Example Limiting Reactant Problem

Combining methane and steam is one way of producing hydrogen gas.

CH4 + H2O → CO + 3H2

a. If 65.0 mol of methane are reacted with 70.0 mol steam, which is the limiting reactant?

b. How many grams of hydrogen are formed? c. How many moles of excess reactant remain?

Limiting Reactant Example Set up for part a/b calculation:

65.0 70.0 ? g

mole mole

CH4 + H2O → CO + 3H2

65.0 mole CH4 x ___________ x ___________ =

gH2

70.0 mole H2O x ___________ x ___________ =

gH2

Butane gas, C4H10, is used in lighters. It burns according to the following reaction.

2C4H10 + 13 O2 → 8CO2 + 10 H2O

a. When 95.0 g of butane is mixed with 250. g of oxygen gas, which is the limiting reactant?

b. What mass in grams of carbon dioxide gas is formed?

c. What mass in grams of excess reactant remains when the reaction is completed?

Example Limiting Reactant Problem

Limiting Reactant ExampleSet up for part a/b calculation:

95.0 g 250.0 g ? g

2C4H10 + 13 O2 → 8CO2 + 10 H2O

95.0 g C4H10 x ___________ x ____________ x _____________

=g CO2

250.0 g O2 x ___________ x ____________ x _____________

=g CO2

Limiting Reactant Example

Set up for part c calculation:

2C4H10 + 13 O2 → 8CO2 + 10 H2O

g _____________x___________x_____________=

g (used in rxn)

95.0 g C4H10 (available at beginning)- g C4H10 (used in reaction)

g C4H10 (excess remaining)

Working a Stoichiometric Problem

First, identify reactants and products and write the balanced equation.

Al + O2 Al2O3

b. What are the reactants?a. Every reaction needs a yield sign!

c. What are the products?

d. What are the balanced coefficients?

4 3 2

6.50 grams of aluminum reacts with 4.23 grams of oxygen to produce aluminum oxide. Identify the limiting and excess reactant.

Working a Stoichiometry Problem Calculating the Limiting

6.50 grams of aluminum reacts with 4.23 grams of oxygen. Identify the limiting and excess reactants.

4Al + 3O2 Al2O3

excess limiting

?g Al2O3 =

?g Al2O3 =

6.50g Al 1 mol Al 2 mol Al2 O3 101.96 g Al2O3 = 12.3

4.23 g O2 1 mole O2 2 mol Al2 O3 101.96 g Al2 O3 = 8.99 g

26.98 g Al 4 mol Al 1 molAl2 O3

g

32.00 g O2 3 mol O2 1 mol Al 2 O3

6.50 g 4.23 g ?g

Based on the past slide, if 6.50 grams of aluminum reacts with 4.23 grams of oxygen calculate the amount of excess that remains.

Always work from the limiting reactant.

Ask yourself how much of the excess was needed or used. ?g needed 4.23 g = l.r.

Al + O2 Al2O34 3 2

?gAl = (needed)

4.23 g O2

32.00 g O2

1 mol O2

3 mol O2

4 mol Al1 mol Al26.98 g Al = 4.76g Al

(needed)

6.50 gram Al placed into the reaction- 4.76 gram Al (needed)

1.74 gram Al wasted = excess remaining

26.98 g 32.00 g

YieldsTheoretical yield – the maximum produced from a balanced chemical equation based on 100% efficiency.

Actual yield – the amount obtained in a laboratory setting. Never 100% due to sources of error.

Percent Yield = actual (lab) x 100

theoretical (math)

#1 - % Yield Example ProblemsCalculate the percentage yield in the following

problem. The theoretical yield of a product is 50.0g

and the actual yield is 41.9 g.

Equation: actual (lab) x 100

theoretical (math)

Answer: 83.8 %

#2 - % Yield Example ProblemsMethyl alcohol, CH3OH, is used as a fuel inalcohol-burning race cars. It can be manufacturedaccording to the following reaction. If 34.5 g of H2

react with excess carbon monoxide, and the actualyield of CH3OH is 252.2 g, what is the % yield ofCH3OH?

2H2 + CO → CH3OH

You need two things to find the percent yield.

Actual yield and theoretical yield! We have the actual, so lets use MATH to find the theoretical!

#3 - % Yield Example ProblemsHydrogen cyanide gas is produced by reacting ammonia gas,

oxygen gas, and methane gas according to the following reaction. If the typical yield is 89.8%, what mass of hydrogen cyanide gas should be expected if 324 g of ammonia gas are reacted with excess oxygen gas and methane gas?

2NH3 + 3O2 + 2CH4 → 2HCN + 6H2O

What does this problem give us? % yield!We can do the math to find the theoretical, but how can we find the actual?

What are we looking for? Actual!

First, lets find the theoretical yield!

?g HCN = 324 g NH3 1 mol NH317.03 g NH3

2 mol HCN2 mol NH3

27.03 g HCN1 mol HCN

The theoretical yield is 514 g HCN.

To find the actual yield, rearrange the % yield formula!

% yield = actual x100theoretical

The rearranged formula is

Actual = % yield x theoretical100

The solution is:

Actual = (89.8) (515 g HCN)100

463 g HCN

#4 - % Yield example problem6.50 grams of aluminum reacts with 4.23 grams of

oxygen. The excess is Al and the limiting is O2. If 7.86 g of Al2O3 are recovered in a lab setting, calculate the percent yield.

4 Al + 3 O2 2Al2O3

First, a theoretical yield must be calculated. A theoretical yields can ONLY be determined from a limiting reactant amount. If only one product is formed, the theoretical yield has already been calculated earlier = 8.99 g Al2O3 ..

Actual yield = 7.86 g

?g Al2O3 =L.R.

4.23 g O2

32.00 g O2

1 mol O2

3 mol O2

2mol Al2O3

1 molAl2O3

101.96 gAl2O3 = 8.99g

% yield = actual yield/theoretical yield x 100 7.86 g / 8.99 g x 100 = 87.4 %