Quantitative Information from Balanced Reactions
The coefficients in a balanced equation represent
• ratios of molecules (formula units)
• ratios of moles
• Use ratios to predict:
- The amount of product formed
- The amount of reactant needed
2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g)
2 molecules C4H10 + 13 molecules O2 
8 molecules CO2 + 10 molecules H2O
2 moles C4H10(l) + 13 moles O2(g) 
8 moles CO2(g) + 10 moles H2O(g)
Use the flowchart below to solve quantitative problems.
2 C4H10(l) + 13 O2(g)  8 CO2(g) + 10 H2O(g)
How many grams of CO2 are formed if 1.00 g of butane
(C4H10) is completely combusted?
Molar masses
C4H10 = 58.123 g/mol
O2
= 31.9988 g/mol
CO2 = 44.010 g/mol
H2O = 18.015 g/mol
Substance “A” on the flowchart is butane.
Substance “B” on the flowchart is carbon dioxide.
1. Convert grams butane to moles butane.
 1molBu tan e 
  0.0172molBu tan e
1.00 gBu tan e

58
.
123
gBu
tan
e


2. Convert moles butane to moles CO2.
 8molCO2 
0.0172molBu tan e
  0.0688molCO2
 2molBu tan e 
3. Convert moles CO2 to grams CO2.
 44.010 gCO2 
  3.03gCO2
0.0688molCO2  
 1molCO2 
For the same combustion reaction, how many grams of
oxygen are required to completely react with 1.00 g of
C4H10?
1. Convert grams butane to moles butane.
(did in previous problem)
0.0172 moles Butane
2. Convert moles butane to moles O2
 13molO2 
0.0172molBu tan e
  0.112molO2
 2molBu tan e 
3. Convert moles O2 to grams O2
 31.9988 gO2 
  3.58 gO2
0.112molO2 

 1molO2 
Limiting Reactants
• If the reactants are not present in stoichiometric
amounts, at end of reaction some reactants are still
present (in excess).
• Limiting Reactant: one reactant that is consumed
Example: 2 H2 + O2 → 2 H2O
Can determine the limiting reactant by making a
comparison of the moles of each reactant.
Compare actual mole ratio to the ratio of the stoichiometric
coefficients.
Example: What is the limiting reactant for the formation of
water from hydrogen and oxygen given 10.98 g of H2 and
27. 45 g of O2?
Molar masses: 1 mol H2 = 2.0158 g H2
1 mol O2 = 31.9988 g O2
1. Convert grams H2 to moles H2
 1molH 2 
  5.447molH 2
10.98 gH 2 

2
.
0158
gH
2 

2. Convert grams O2 to moles O2
 1molO2 
  0.8578molO2
27.45 gO2 
31
.
9988
gO
2 

3. Find mole ratio and compare to ratio of the
stoichiometric coefficients from the balanced reaction.
Be sure to be consistent with which substance is in the
numerator and which is in the denominator.
Actual ratio:
5.447molH 2
6
0.8578molO2
Coeff.
2molH 2
Ratio: 1molO  2
2
O2 is the limiting reagent and will be used up.
H2 is in excess and some will be left over.
When comparing the ratios as above:
-if the actual ratio of moles < coefficient ratio
The substance in the numerator is the limiting reactant.
- if the actual ratio of moles > coefficient ratio
The substance in the denominator is the limiting reactant.
Theoretical Yields
• The maximum amount of product predicted from
stoichiometry taking into account limiting reagents is
called the theoretical yield.
• The limiting reactant determines the maximum amount
of product that can be produced (theoretical yield).
• Limiting reactant gets all used up. Other reactants are
in excess.
What is the theoretical yield of water for the given
amounts of hydrogen and oxygen above?
1. Find moles of O2 (did already) = 0.8578 mol O2
2. Find moles water from moles O2
 2molH 2 O 
  1.715molH 2 O
0.8578molO2 

 1molO2 
3. Find grams water from moles water
 18.0152 gH 2 O 
  30.90 gH 2 O
1.715molH 2 O

 1molH 2 O 
Also give the amount (in grams) of the reactant that is left
in excess.
1. Find moles O2 (did already) = 0.8578 mol O2
2. Find moles of H2 used (reacted) from moles of O2
 2molH 2 
  1.715molH 2
0.8578molO2 

 1molO2 
3. Find grams of H2 (used up) from moles H2
 2.0158 gH 2 
  3.457 gH 2
1.715molH 2 
 1molH 2 
4. Find unused (excess) moles of H2 from original
mass (10.98 g) of H2
10.98 g H2 – 3.457 g H2 = 7.523 g H2 left over
Percent Yield
• The percent yield relates the actual yield (amount of material
recovered in the laboratory) to the theoretical yield:
% Yield 
Actual yield
100
Theoretica l yield
What is the percent yield of water, given the amounts of hydrogen and
oxygen above, if the actual yield is 19.64 g of water?
19.64 gO2
% yield 
100  63.56%
30.90 gO2