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
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