1 ib topic 1: quantitative chemistry 1.4: mass relationships in chemical reactions 1.4.1 calculate...

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IB Topic 1: Quantitative Chemistry1.4: Mass Relationships in Chemical Reactions

1.4.1 Calculate theoretical yields from chemical equations.

1.4.2 Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given.

1.4.3 Solve problems involving theoretical, experimental and percentage yield.

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Review: Identify the mole ratio of any two species in a chemical reaction.

How do you calculate the quantity of reactants and products in chemical reactions?

Use the chemical equation to predict ratio of reactants and productsN2 + 3H2 2NH3

1 molecule N2 reacts with 3 molecules H2 producing 2 molecules NH3

1 mole N2 reacts with 3 moles H2 producing 2 moles NH3

The balanced equations gives us the ratio in particles or moles (not mass) of the chemicals involved in the reaction.The ratio of N2 to NH3 is 1:2The ratio of NH3 to H2 is 2:3The ratio of H2 to N2 is 3:1

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Review: Identify the mole ratio of any two species in a chemical reaction.

Consider the reaction:

2C8H18 + 25O2 16CO2 + 18H2O

What is the ratio of O2 to CO2? 25:16

What is the ratio of H2O to C8H18? 18:2

What is the ratio of CO2 to H2O?

What is the ratio of O2 to C8H18?

Remember these ratios are in particles or moles. We will be using moles.

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• Theoretical Yield is the amount of product that would result if all the limiting reagent reacted.

• In essence, theoretical yield is what you calculate

• We use a process called stoichiometry: It is the calculation of quantities in chemical reactions.

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How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal?

2 Na + Cl2 2 NaCl2.6 moles Cl2 2 mol NaCl

1 mol Cl2= 5.2 moles NaCl

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1.4.1 Calculate theoretical yields from chemical equations. Most of the time in chemistry, the amounts are given

in grams instead of moles We still go through moles and use the mole ratio, but

now we also use molar mass to get to grams Example: How many grams of chlorine are required to

react completely with 5.00 moles of sodium to produce sodium chloride?

2 Na + Cl2 2 NaCl

5.00 moles Na 1 mol Cl2 70.90g Cl2

2 mol Na 1 mol Cl2

= 177g Cl2

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Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum.

2 Al + 3 I2 2 AlI3

0.50 moles Al 3 moles I2 253.8g I2

2 moles Al 1 mole I2

= 190.35 g I2

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We can also start with mass and convert to moles of product or another reactant

We use molar mass and the mole ratio to get to moles of the compound of interest Calculate the number of moles of ethane (C2H6) needed

to produce 10.0 g of water 2 C2H6 + 7 O2 4 CO2 + 6 H20

10.0 g H2O 1 mol H2O 2 mol C2H6

18.0 g H2O 6 mol H20

= 0.185 mol C2H6

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Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide

4 Al + 3 O2 2 Al2O3

10.0g Al10.0g Al22OO3 3 1 mol Al 1 mol Al22OO3 3 3 mol O3 mol O2 2

101.96g Al101.96g Al22OO3 3 2 mol Al 2 mol Al22OO33

= 0.147 mol O= 0.147 mol O22

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1.4.1 Calculate theoretical yields from chemical equations. Most often we are given a starting mass and

want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!)

Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in

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Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen.

N2 + 3 H2 2 NH3

2.00g N2 1 mol N2 2 mol NH3 17.06g NH3

28.02g N2 1 mol N2 1 mol NH3

= 2.4 g NH3

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How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen?

3 Ca + N2 Ca3N2

2.00g Ca 1 mol Ca 1 mol CaCa33NN2 2 148.26g Ca 148.26g Ca33NN2 2

40.08g Ca 3 mol Ca 1 mol 40.08g Ca 3 mol Ca 1 mol CaCa33NN22

= 2.47 g CaCa33NN22

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How much oxygen does it take to burn 100.0 grams of octane?

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Write the balanced chemical equation:2C8H18 + 25O2 16CO2 + 18H2O

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2C8H18 + 25O2 16CO2 + 18H2O

Mass of O2 required: 10.97 mol (32.00 g/mol) =351.0 g O2

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In a spectacular reaction called the thermite reaction, iron(III) oxide reacts with aluminum producing iron and aluminum oxide. How many grams of iron will be produced from 43.7

grams of aluminum?

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grams of aluminum?

Write the balanced chemical equation: Fe2O3 + 2Al 2Fe + Al2O3

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grams of aluminum?

Write the balanced chemical equation: Fe2O3 + 2Al 2Fe + Al2O3

Mass of Fe produced: 43.7 grams Al x 1 mole Al/ 26.98 Al x 2Fe/2Al x 55.8 g/ 1mol) = 90.4

g Fe

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Limiting Reactant: Cookies1 cup butter

1/2 cup white sugar

1 cup packed brown sugar

1 teaspoon vanilla extract

2 eggs

2 1/2 cups all-purpose flour

1 teaspoon baking soda

1 teaspoon salt

2 cups semisweet chocolate chips

Makes 3 dozen

If we had the specified amount of all ingredients listed, could we make 4 dozen cookies?

What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies?

What if we only had one egg, could we make 3 dozen cookies?

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1.4.2 Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given.

Limiting Reactants

You are given amounts for two reactants and one reactant will run out first. This is called the limiting reactant. The reactant that is left over is called the excess reactant.

For example: A strip of zinc metal weighing 2.00 g is placed in a solution containing 2.50 g of silver nitrate causing the following reaction to occur:

Zn + 2AgNO3 2Ag + Zn(NO3)2

How many grams of Ag will be produced?

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Limiting Reactant Most of the time in chemistry we have more of one

reactant than we need to completely use up other reactant.

That reactant is said to be in excess (there is too much).

The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant.

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Method for determining the L.R.1) Pick A Product

2) Try ALL the reactants

3) The lowest answer will be the correct answer

4) The reactant that gives the lowest answer will be the limiting reactant

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Limiting Reactant 10.0g of aluminum reacts with 35.0 grams of chlorine

gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced?

2 Al + 3 Cl2 2 AlCl3

Start with Al:

Now Cl2:

10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g AlCl3

27.0 g Al 2 mol Al 1 mol AlCl3

= 49.4g AlCl3

35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3

71.0 g Cl2 3 mol Cl2 1 mol AlCl3

= 43.9g AlCl3

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LR Example Continued We get 49.4g of aluminum chloride from the

given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete .

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Limiting Reactant Practice 15.0 g of potassium reacts with 15.0 g of

iodine. Calculate which reactant is limiting and how much product is made.

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Limiting Reactant Practice2 K + I2 2 KI Potassium:

Iodine:

Iodine is the limiting reactant and we get 19.6 g of potassium iodide

15.0g K 1 mol K 2 mol KI 166 g KI

39.1g K 2 mol K 1 mol KI= 63.7g KI

15.0 g I2 1 mol I2 2 mol KI 166 g KI

254 g I2 1 mol I2 1 mol KI= 19.6 g KI

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Finding the Amount of Excess By calculating the amount of the excess

reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.

Can we find the amount of excess potassium in the previous problem?

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Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine.

2 K + I2 2 KI We found that Iodine is the limiting reactant, and

19.6 g of potassium iodide are produced.

15.0 g I2 1 mol I2 2 mol K 39.1 g K

254 g I2 1 mol I2 1 mol K= 4.62 g K USED!

15.0 g K – 4.62 g K = 10.38 g K EXCESS

Given amount of excess reactant

Amount of excess reactant actually used

Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

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Limiting Reactant: Recap1. You can recognize a limiting reactant problem because

there is MORE THAN ONE GIVEN AMOUNT.2. Convert ALL of the reactants to the SAME product

(pick any product you choose.)3. The lowest answer is the correct answer.4. The reactant that gave you the lowest answer is the

LIMITING REACTANT.5. The other reactant(s) are in EXCESS.6. To find the amount of excess, subtract the amount used

from the given amount.7. If you have to find more than one product, be sure to

start with the limiting reactant. You don’t have to determine which is the LR over and over again!

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Putting it all together At high temperatures, sulfur combines with

iron to form the brown-black iron (II) sulfide: Fe (s) + S (l) FeS (s) In one experiment, 7.62 g of Fe are allowed

to react with 8.67 g of S. What is the limiting reagent, and what is the

reactant in excess? How much of the excess reactant is left over?

Calculate the mass of FeS formed.

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Putting it all together Answers Fe is the limiting reagent S is in excess ??? g of S is left over 12.2 g FeS formed

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The theoretical yield from a chemical reaction is the yield calculated by assuming the reaction goes to completion.

In practice we often do not obtain as much product from a reaction mixture as theoretically possible due to a number of factors:

Many reactions do not go to completion Some reactants may undergo two or more reactions simultaneously forming

undesired products Not all of the desired product can be separated from the rest of the productsThe amount of a specified pure product actually obtained from a given reaction is

the experimental or actual yield.

Percentage yield indicates how much of a desired product is obtained from a reaction

Percentage yield = experimental yield of a product x 100% theoretical yield of a product

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Consider the reaction: P4 + 5O2 P4O10

If 272 g of phosphorus reacts, the percentage yield of tetraphosphorus decoxide is 89.5%. What mass of P4O10 is obtained?

Theoretical amt: 272 g/1 x 1mole/124.0 g x 1mol P4O10/ 1mole P4 x 284 g/

1mol =622.9 g

Experimental amt: 623 g x .895 = 558 g

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Consider the reaction: P4O10 + 6H2O 4H3PO4

If 558 g of P4O10 reacts, the experimental yield of H3PO4 is 746 g. What is the percentage yield of H3PO4?

Theoretical amt: 558 g/1 x 1 mol/284 g x 4 mol H3PO4 / 1 mol P4O10 x 98.0 g / 1

mol = X

X= 770. g H3PO4

Percentage yield: 746 g x 100% = 96.9%

770 g

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Consider the reaction: Fe2O3 + 3CO 2Fe + 3CO2

When 84.8 g iron(III) oxide reacts with an excess of carbon monoxide, 54.3 g of iron is produced. What is the percentage yield of this reaction?

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Consider the reaction: Fe2O3 + 3CO 2Fe + 3CO2

When 84.8 g iron(III) oxide reacts with an excess of carbon monoxide, 54.3 g of iron is produced. What is the percentage yield of this reaction?

Theoretical amt: 84.8 g/1 x 1 mol/159.2 g x 2 mol Fe / 1 mole Fe2O3 x 55.8 g/ 1mol

Fe2O3 = X

X = 59.4 g FePercentage yield: 54.3 g x 100% = 91.4%

59.4 g

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If 50.0 g silicon dioxide is heated with an excess of carbon, the percentage yield of silicon carbide 76.0%. What mass of silicon carbide will actually be produced? SiO2 + 3C SiC + 2CO

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If 50.0 g silicon dioxide is heated with an excess of carbon, the percentage yield of silicon carbide 76.0%. What mass of silicon carbide will actually be produced? SiO2 + 3C SiC + 2CO

Theoretical amt: 50.0 g/ 1 x 1 mol/60.1 g x 1mol SiC/1mol SiO2 x 40.1 g/

1mol =X

X = 33.4 g SiC

Experimental yield: 33.4 g x .760% = 25.4 g SiC

1 ib topic 1: quantitative chemistry 1.4: mass relationships in chemical reactions 1.4.1 calculate...

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