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Chapter 12 Stoichiometry

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Page 1: Chapter 12 Stoichiometry. Stoichiometry Topics Conservation of mass and the balanced chemical equation Mole ratios in a chemical equation Stoichiometric

Chapter 12

Stoichiometry

Page 2: Chapter 12 Stoichiometry. Stoichiometry Topics Conservation of mass and the balanced chemical equation Mole ratios in a chemical equation Stoichiometric

Stoichiometry Topics

Conservation of mass and the balanced chemical equation

Mole ratios in a chemical equationStoichiometric calculations involving

reactants and productsLimiting ReactantsPercent Yield

Page 3: Chapter 12 Stoichiometry. Stoichiometry Topics Conservation of mass and the balanced chemical equation Mole ratios in a chemical equation Stoichiometric

What is Stoichiometry?

Stoichiometry is the study of quantitative relationships between amounts of reactants used and products formed by a chemical reaction.

Page 4: Chapter 12 Stoichiometry. Stoichiometry Topics Conservation of mass and the balanced chemical equation Mole ratios in a chemical equation Stoichiometric

Practical Applications of Stoichiometry

In a spacecraft, the carbon dioxide exhaled by astronauts can be removed by its reaction with lithium hydroxide, LiOH, according to the following chemical equation.

CO2(g) + 2LiOH(s) Li2CO3(s) + H2O(l)

How many moles of lithium hydroxide are required to react with 20 mol of CO2, the average amount exhaled by a person each day?

Page 5: Chapter 12 Stoichiometry. Stoichiometry Topics Conservation of mass and the balanced chemical equation Mole ratios in a chemical equation Stoichiometric

Strategy – Relate everything to the mole

The mole is the common thread that links reactants to products.

1CO2(g) + 2LiOH(s) 1Li2CO3(s) + 1H2O(l)

Mole Ratios

molLiOH

molCO

2

1 2

32

2

1

1

COmolLi

molCOOmolH

molCO

2

2

1

1

321

2

COmolLi

molLiOHOmolH

molLiOH

21

2

Page 6: Chapter 12 Stoichiometry. Stoichiometry Topics Conservation of mass and the balanced chemical equation Mole ratios in a chemical equation Stoichiometric

The Mole Super Highway

Mole

Mass

Particles

1 mole = molar mass

1 mole = 6.02 x 1023 particles

Mole

Mass

Particles

1 mole = molar mass

1 mole = 6.02 x 1023 particles

MOLE RATIO1 mole = 22.4L at (STP) 1 mole = 22.4L at (STP)

VolumeVolume

Page 7: Chapter 12 Stoichiometry. Stoichiometry Topics Conservation of mass and the balanced chemical equation Mole ratios in a chemical equation Stoichiometric

What mass of oxygen is needed to fully combust 12.5g ethane (C2H6)?

OHCOOHC 22262 ________

Practice ProblemsHow many moles of water are formed when 3.8 moles of Hydrogen react with an excess of Oxygen?

Page 8: Chapter 12 Stoichiometry. Stoichiometry Topics Conservation of mass and the balanced chemical equation Mole ratios in a chemical equation Stoichiometric

Practice Problem Involving Volume (not at STP)

What is the volume of nitrogen gas that would be produced from 12.6g of ammonia (NH3)?

The density of nitrogen is 0.808g/ml under the temperature conditions for this experiment.

Page 9: Chapter 12 Stoichiometry. Stoichiometry Topics Conservation of mass and the balanced chemical equation Mole ratios in a chemical equation Stoichiometric

Limiting Reactants

Why do reactions end?Some or all of the reactants get used up.

Limiting Reactants limit the extent of the reaction and thereby determines the amount of the product.

Leftover reactants are called excess reactants

Page 10: Chapter 12 Stoichiometry. Stoichiometry Topics Conservation of mass and the balanced chemical equation Mole ratios in a chemical equation Stoichiometric

How do you determine which reactant is limited?

If 200g of sulfur reacts with 100g of chlorine, what mass of disulfur dichloride is produced?

1. Determine the moles of the reactants based on their masses.

2. Determine if the two reactants are in the correct mole ratio as given by the balanced chemical equation.

3. Use the number of moles of the limiting reaction as your given value to calculate the mass of the product that could be produced.

2228 44 ClSClS

Page 11: Chapter 12 Stoichiometry. Stoichiometry Topics Conservation of mass and the balanced chemical equation Mole ratios in a chemical equation Stoichiometric

What else can you do ….

Determine how much reactant is in excess once the reaction is carried out 1. Calculate the actual amount of the excess reactant that is required by using the given quantity of the limiting reactant

2. Calculate the difference between the amount of the excess reactant required and the amount that is given.

Page 12: Chapter 12 Stoichiometry. Stoichiometry Topics Conservation of mass and the balanced chemical equation Mole ratios in a chemical equation Stoichiometric

Practice Problem

Chlorine is used by textile manufacturers to bleach cloth. Excess chlorine is destroyed by its reaction with sodium thiosulfate, :__Na2S2O3(aq) + __Cl2(g) + __H2O(aq) __NaHSO4(aq) + __HCl(aq)

 What is the mole ratio Na2S2O3 to Cl2 in the balanced chemical equation?

How many grams sodium thiosulfate is needed to completely destroy 103g of chlorine?   Is 75g of Na2S2O3 an adequate quantity to destroy 35g of Cl2 ? 

 Which reactant is in excess? And how much is left over after the reaction occurs?

Page 13: Chapter 12 Stoichiometry. Stoichiometry Topics Conservation of mass and the balanced chemical equation Mole ratios in a chemical equation Stoichiometric

Percent Yield

Stoichiometric calculations provide a theoretical yield or maximum amount of product that can be produced from a given amount of reactants.

The actual yield is the amount that is actually produced when a reaction is carried out.

The percent yield reflects the ratio of the theoretical and actual yields

100)(

)(x

tryStoichiomelYieldTheoretica

ExperimentdActualYielldPercentYie

Page 14: Chapter 12 Stoichiometry. Stoichiometry Topics Conservation of mass and the balanced chemical equation Mole ratios in a chemical equation Stoichiometric

Percent Yield Problem

Ethanol (C2H5OH) is produced from the fermentation of sucrose.

Determine the theoretical and percent yields of ethanol if 684g sucrose undergoes fermentation and 349g of ethanol are obtained.

2522112212 440 COOHHCHOHC