StoichiometryMass Changes in Chemical Reactions

Limiting reactantsPercentage yield

Stoichiometry ProblemsHow many moles of KClO3 must

decompose in order to produce 9 moles of oxygen gas?

= 6 mol KClO3

Problem: X molKClO3 9mol O2 2KClO3 2KCl + 3O2 Balanced : 2 molKClO3 3mol O2

Equation

3molO2

29

molKClO3 2

XmolKClO3 molO

Example 1

Stoichiometry ProblemsHow many grams of silver will be formed

when 12 g of copper reacts with aluminum nitrate to produce silver and copper II nitrate and silver?

= 41 g Ag

Problem: 12gCu Xg Ag

Cu + 2 AgNO3 2 Ag + Cu(NO3)2

Balanced: 63.5 gCu 2(107.9) g AgEquation 215.8 g

63.5gCu

12

215.8gAg

XgAg gCu

Example2

Stoichiometry ProblemsIf 12.0 grams of potassium chlorate

decompose, how many moles of potassium chloride will be produced?

= 0.0988 mol KCl

Problem: 12gKClO3 X moles KCl

2 KClO3 2KCl + 3 O2

Balanced: 2(122.6) g KClO3 2 moles KClEquation 245.2 g

gKClO3 245.2

312

molKCl 2

KCl Xmol gKClO

Example3

Stoichiometry Problems In an experiment, red mercury (II) oxide powder is placed in

an open flask and heated until it is converted to liquid mercury and oxygen gas. The liquid mercury has a mass of 92.6 g. What is the mass of oxygen formed in the reaction?

= 7.39 g O2

Problem: 92.6 g Hg + X g O2

2HgO 2Hg + O2

Balanced: 2 ( 200.6) g Hg + 32 g O2

Equation 401.2 g Hg

Hg g 401.2

6.92

O2 32g

O2 Xg gHg

LEARNINGCHECK

Limiting Reactant

Bike AnalogyConsider the following Analogy:

2 Wheels + 1 Body + 1 Handle bar + 1 Gear Chain = 1 bike

How many bikes can be produced given, 8 wheels, 5 bodies, 6 handle bar and 5 gear chain?

Bike Analogy

Limiting Reactant

Excess Reactant

Consider the following Analogy:

2 Wheels + 1 Body + 1 Handle bar + 1 Gear Chain = 1 bike

How many bikes can be produced given, 8 wheels, 5 bodies, 6 handle bar and 5 gear chain?

Cheeseburger AnalogyConsider the following Analogy:2 Cheese + 1 burger patty + 1 bun

= cheese burger

Given three hamburger patties, six buns, and 12 slices of cheese, how many cheese burger can be made?

Cheeseburger Analogy

Consider the following Analogy:2 Cheese + 1 burger patty + 1 bun

= 1 cheese burger

Given three hamburger patties, six buns, and 12 slices of cheese, how many cheese burger can be made?

LR

ER

Limiting Reactant vs. Excess Reactants

–Limiting reactant is the reactant that runs out first

–When the limiting reactant is exhausted, then the reaction stops

In our examples, the limiting reactants will be the wheels in the bicycle analogy and the burger patty in our hamburger analogy

1. Write a balanced equation.

2. For each reactant, calculate the amount of

product formed.

3. The reactant that resulted in the smallest amount of

product is the limiting reactant(LR).

4. To find the amount of leftover reactant—excess—

calculate the amount of the no LR used by the LR.

5. Subtract the calculated amount in step 4 from the

original no LR amount given in the problem.

Limiting Reactants Calculations

Example 1 Determine how many moles of water can be

formed if I start with 2.75 moles of hydrogen and 1.75 moles of oxygen.

= 2.75 mol H2O

Problem: 2.75 mol H2 XmolH2O

2H2 + O2 2H2O Balanced: 2 mol H2 2molH2OEquation

H2 mol 2

H2 mol 2.75

2molH2O

XmolH2O

= 3.50 mol H2O

Problem: 1.75 mol O2 XmolH2O

2H2 + O2 2H2O Balanced: 1 mol O2 2molH2OEquation

O2 mol 1

O2 mol 1.75

2molH2O

XmolH2O

Limiting reactant =H2

If 2.0 mol of HF are exposed to 4.5 mol of SiO2, which is the limiting reactant?

= 1.0 mol H2O

Problem: 2.0 mol HF XmolH2OSiO2(s) + 4HF(g) SiF4(g) +

2H2O(l)Balanced: 4 mol HF 2molH2OEquation

HF 4.0mol

HF 2.0mol

molH2O 2

XmolH2O

= 9.0 mol H2O

Problem: 4.5 mol SiO2 XmolH2OSiO2(s) + 4HF(g) SiF4(g) +

2H2O(l)Balanced: 1 mol SiO2 2molH2OEquation

1.0molSiO2

4.5molSiO2

molH2O 2

XmolH2O

Limiting reactant =HF

Example2

If 36.0 g of H2O is mixed with 167 g of Fe , which is the limiting reactant?

= 106 g Fe2O3

Problem: 36.0 g H2O XgFe2O3

2Fe(s) + 3H2O(g) Fe2O3(g) + 3H2(g)Balanced: 54 g H2O 159.6gFe2O3

Equation

54.0gH2O

H2O 36.0g

3159.6gFe2O

XgFe2O3

Limiting reactant =H2O

= 238 g Fe2O3

Problem: 167 g Fe XgFe2O3

2Fe(s) + 3H2O(g) Fe2O3(g) + 3H2(g)Balanced: 111.6 g Fe 159.6gFe2O3

Equation

gFe 111.6

Fe 167g

3159.6gFe2O

XgFe2O3

LEARNINGCHECK

1. Write a balanced equation.

2. For each reactant, calculate the amount of

product formed.

3. The reactant that resulted in the smallest amount of

product is the limiting reactant(LR).

4. To find the amount of leftover reactant—excess—

calculate the amount of the no LR used by the LR.

5. Subtract the calculated amount in step 4 from the

original no LR amount given in the problem.

Limiting Reactants Calculations

Limiting Reactants

Limiting reactant: H2O

Excess reactant: Fe

Products Formed: 107 g Fe3O3 & 4.00 g H2

left over iron

3Fe(s) + 4H2O(g) Fe3O3(g) + 4H2(g)

LRXS

3Fe(s) + 4H2O(g) Fe3O3(g) + 4H2(g)

Problem: XgFe 36.0 g H2O

Balanced: 111.6 g Fe 54 g H2OEquation

gH2O 54

36gH2O

111.6gFe

XgFe = 74.4 g Fe used

167gFe - 74.4 g Fe= 92.6 g FeOriginal – Used = Excess

So far, the masses we have calculated from chemical equations were based on the assumption that each reaction occurred 100%.

The THEORETICAL YIELD the maximum amount of product that can be produced in a reaction (calculated from the balanced equation)

The ACTUAL YIELD is the amount of product that is “actually” produced in an experiment (usually less than the theoretical yield)

Percent Yield

Percent Yield

Theoretical Yield the maximum amount of product that

can be produced in a reactionPercent Yield

◦The actual amount of a given product as the percentage of the theoretical yield.

Look back at the problem from LEARNING CHECK. We found that 106 g Fe2O3 could be formed from the reactants.

In an experiment, you formed 90.4 g of Fe2O3. What is your percent yield?

% Yield = 90.4 g x 100 = 85.3% 106 g

A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate, CH3COOC2H5. What percent yield of ethyl acetate is this?

= 19.1 g CH3COOC2H5

Problem: 10.0g C2H5OH Xg CH3COOC2H5

CH3COOH + C2H5OH CH3COOC2H5 + H2O

Balanced: 46.0 g C2H5OH 88.0 g CH3COOC2H5

Equation

H46.0gC2H5O

C2H5OH0.10

OC2H588.0gCH3CO

H5XgCH3COOC2 g

% Yield = 14.8 g x 100 = 77.5% 19.1g

Example1

When 36.8 g of C6H6 reacts with Cl2, what is thetheoretical yield of C6H5Cl produced? If the actual is43.7 g, determine the percentage yield of C6H5Cl.

= 53.1 g C6H5Cl

Problem: 36.8g C5H5 Xg C5H5Cl

2C6H6 + Cl2 2C6H5Cl + H2

Balanced: 156.0 g C5H5 225.0 g C5H5Cl

Equation

660.156

668.36

Cl225.0gC6H5

XgC6H5Cl

HgC

HgC

% Yield = 43.7 g x 100 = 88.3% 53.1g

LEARNINGCHECK