UNIT 4
Quantitative Information from
Balanced Chemical Equations
Quantitative Information from Balanced
Equations
Getting information from a balanced chemical equation is a
matter of putting together your knowledge of chemical
formulas and mass/moles/numbers relationships, and then
applying them to a balanced chemical reaction.
Example: Propane is a common fuel. How many moles of
oxygen are needed to burn 3.62 moles of propane?
1. Write and balance the equation for the reaction:
3.62 mol ? mol
C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l)
2. Write what you know. Write what you need to know.
Quantitative Information from Balanced
Equations
3.62 mol ? mol
C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l)
3. Convert your “knowns” to moles, if necessary.
4. Use the appropriate stoichiometric equivalences from the
equation to convert from moles of what you have to moles of
what you need.
5. Convert your answer from moles to the required units.
3.62 mol C3H8 x
5 mol O2 =
1 mol C3H8
18.1 mol O2
stoichiometric equivalence
(from chemical equation)
Here we were asked for moles, so no mole/mass conversion was necessary.
Quantitative Information from Balanced
Equations
Example: What mass of oxygen is consumed in the combustion
of 100.0 g of propane?
1. Write and balance the equation for the reaction.
2.
3.
4.
5.
100.0 g
?g
C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l)
Write what you know and what you need to know.
Convert “what you know” to moles.
Using the stoichiometry of the reaction, convert from moles of “what you
know” to moles of “what you need to know.”
Convert the moles of what you need to appropriate final units.
100.0 g C3H8 x 1
mol C3H8 x 5 mol O2 x 32.00 g O2 =
44.10 g
1 mol C3H8 1 mol O2
mass  moles
stoichiometric equivalence
362.8 g O2
moles  mass
Quantitative Information from Balanced
Equations - Example
Detonation of nitroglycerin proceeds as follows:
4C3H5N3O9(l)  12CO2(g) + 6N2(g) + O2(g) + 10H2O(g)
a) If a sample containing 2.00 mL of nitroglycerin (density
= 1.592 g/mL) is detonated, how many total moles of
gas are produced?
b) If each mole of gas occupies 55 L under the conditions
of the explosion, how many liters of gas are produced?
c) How many grams of N2 are produced in the detonation?
answers: a) 0.102 mol, b) 5.6 L, c) 0.589 g
Limiting Reactant
N2 (g) + 3H2 (g)  2NH3 (g)
If we start with 2 moles nitrogen and 6 moles hydrogen, how
many moles of ammonia is it theoretically possible to
produce?
One way to work the problem:
A second way:
6 mol H2 x
2 mol N2 x
2 mol NH3
3 mol H2
=
2 mol NH3
1 mol N2
=
4 mol NH3
4 mol NH3
4 moles of ammonia is all the ammonia we can get when we react
2 moles nitrogen and 6 moles hydrogen. This amount of
product is called the theoretical yield.
Limiting Reactant
N2 (g) + 3H2 (g)  2NH3 (g)
If we start with 2 moles nitrogen and 7 moles hydrogen, how
many moles of ammonia is it theoretically possible to
produce?
Still just 4 moles…we do not have enough N2 to make more
than 4 moles of NH3. We call the reactant that is NOT in
excess the limiting reactant.
In this example, N2 is the limiting reactant. H2 is in excess.
Limiting reactant problems are usually ones in which amounts of
all reactants are given (as opposed to just one).
Limiting Reactant
Example: A chemist takes 20.00 g KAuCl4 and 25.00 g Na2CO3 and dissolves
both in an excess of water. What is the theoretical yield of Au(OH)3, if
the equation for the reaction is the one given below?
2KAuCl4(aq) + 3Na2CO3(aq) + 3H2O(l)  2Au(OH)3(s) +
6NaCl(aq) + 2KCl(aq) + 3CO2(g)
This problem is worked with the same five steps as for the
stoichiometry equations, BUT…
you must perform the calculation for each reactant (unless
it’s reported to be in excess)!
Whichever reactant gives you LESS product is the limiting
reactant.
Limiting Reactant
Example: A chemist takes 20.00 g KAuCl4 and 25.00 g Na2CO3 and dissolves
both in an excess of water. What is the theoretical yield of Au(OH)3, if
the equation for the reaction is the one given below?
20.00 g
25.00 g
?g
2KAuCl4(aq) + 3Na2CO3(aq) + 3H2O(l)  2Au(OH)3(s) +
6NaCl(aq) + 2KCl(aq) + 3CO2(g)
Calculate the theoretical yield of Au(OH)3 from 20.00 g KAuCl4:
20.00 g KAuCl4 x 1 mol KAuCl4 x 2 mol Au(OH)3 x 247.99 g Au(OH)3
377.88 g KAuCl4
2 mol KAuCl4
1 mol Au(OH)3
= 13.13 g Au(OH)3
The results of this calculation mean that if we had 20.00 g KAuCl4
and an excess of Na2CO3, we could produce 13.13 g of
Au(OH)3.
Limiting Reactant
Example: A chemist takes 20.00 g KAuCl4 and 25.00 g Na2CO3 and dissolves
both in an excess of water. What is the theoretical yield of Au(OH)3, if
the equation for the reaction is the one given below?
20.00 g
25.00 g
?g
2KAuCl4(aq) + 3Na2CO3(aq) + 3H2O(l)  2Au(OH)3(s) +
6NaCl(aq) + 2KCl(aq) + 3CO2(g)
Calculate the theoretical yield of Au(OH)3 from 25.00 g Na2CO3:
25.00 g Na2CO3 x 1 mol Na2CO3 x 2 mol Au(OH)3 x 247.99 g Au(OH)3
105.99 g Na2CO3
3 mol Na2CO3
1 mol Au(OH)3
= 39.00 g Au(OH)3
The results of this calculation mean that if we had 25.00 g Na2CO3
and an excess of KAuCl4, we could produce 39.00 g of
Au(OH)3.
Limiting Reactant
Example: A chemist takes 20.00 g KAuCl4 and 25.00 g Na2CO3 and dissolves
both in an excess of water. What is the theoretical yield of Au(OH)3, if
the equation for the reaction is the one given below?
20.00 g
25.00 g
?g
2KAuCl4(aq) + 3Na2CO3(aq) + 3H2O(l)  2Au(OH)3(s) +
6NaCl(aq) + 2KCl(aq) + 3CO2(g)

If we had 20.00 g KAuCl4 and an excess of Na2CO3, we could
produce 13.13 g of Au(OH)3.

If we had 25.00 g Na2CO3 and an excess of KAuCl4, we could
produce 39.00 g of Au(OH)3.

13.13 g is the smaller amount of product, which means
KAuCl4 is the limiting reactant, and the theoretical yield
is 13.13 g.
Limiting Reactant
Example: A chemist takes 20.00 g KAuCl4 and 25.00 g Na2CO3 and dissolves
both in an excess of water. How much of the excess reactant is present
at the completion of the reaction?
20.00 g
?g
13.12 g
2KAuCl4(aq) + 3Na2CO3(aq) + 3H2O(l)  2Au(OH)3(s) +
6NaCl(aq) + 2KCl(aq) + 3CO2(g)
First method: Use the limiting reactant to calculate how much of
the excess reactant is actually needed in the reaction.
20.00 g KAuCl4 x 1 mol KAuCl4 x 3 mol Na2CO3 x 105.99 g Na2CO3
377.88 g KAuCl4 2 mol KAuCl4
1 mol Na2CO3
= 8.415 g Na2CO3
The results of this calculation mean 8.415 g Na2CO3 is the mass
needed to react with 20.00 g KAuCl4. Any mass beyond that
is excess:
Excess Na2CO3 = 25.00 g – 8.415 g = 16.58 g
Limiting Reactant
Example: A chemist takes 20.00 g KAuCl4 and 25.00 g Na2CO3 and dissolves
both in an excess of water. How much of the excess reactant is present
at the completion of the reaction?
20.00 g
?g
13.12 g
2KAuCl4(aq) + 3Na2CO3(aq) + 3H2O(l)  2Au(OH)3(s) +
6NaCl(aq) + 2KCl(aq) + 3CO2(g)
Second method: Use the theoretical yield to calculate how much
of the excess reactant is actually needed in the reaction.
13.13 g Au(OH)3 x 1 mol Au(OH)3 x 3 mol Na2CO3 x 105.99 g Na2CO3
247.99 g Au(OH)3 2 mol Au(OH)3
1 mol Na2CO3
= 8.418 g Na2CO3
The results of this calculation mean 8.418 g Na2CO3 is the mass
needed to produce 13.13 g Au(OH)3. Any mass beyond that
is excess:
Excess Na2CO3 = 25.00 g – 8.418 g = 16.58 g
Limiting Reactant – Percent Yield
Often reactions do not go to 100% completion. Percent
yield gives a quantitative measure of the extent to
which the reaction went to completion.
Percent yield
=
100 x
actual yield
theoretical yield
Example: A chemist takes 20.00 g KAuCl4 and 25.00 g Na2CO3 and dissolves
both in an excess of water. The Au(OH)3 recovered is found to have a
mass of 10.83 g. What is the percent yield of the reaction?
We already have calculated the theoretical yield and found it to be 13.13g.
% yield = 100 x 10.83 g = 82.48 %
13.13 g
One more limiting reactant problem
The reaction below is run starting with 1.000 g
nitrogen and 2.000 g lithium. Report the
grams of each species present at the end of
the reaction if a) the yield is 100% and b)
the yield is 75%.
6Li(s) + N2 (g)  2Li3N(s)
100%: Li = 0.513 g, N2 = 0 g, Li3N = 2.487 g
75%: Li = 0.885 g, N2 = 0.250 g, Li3N = 1.865 g
More problems
How many kilograms of aluminum are
required to produce 2.00 kilograms of
hydrogen according to the following
reaction?
2Al(s) + 6H+(aq)  3H2(g) + 2Al3+(aq)
17.8 kg
More problems
One molecule of penicillin G has a mass of
5.553 x 10-22 g. What is the molar mass of
penicillin G?
334.4 g
Hemoglobin has four iron atoms per molecule
and contains 0.340% iron by mass.
Calculate the molar mass of hemoglobin.
65700 g
More problems
Serotonin has 68.2 wt% C, 6.86 wt% H, 15.9
wt% N, and 9.08 wt% O. Its molar mass is
176 g. Write its molecular formula.
C10H12N2O
Name or write formula
HNO2
H2SO4
HF(aq)
HC2H3O2
FeP
CuOH
K2CO3
NaHCO3
NH3
permanganic acid
iron(II) sulfate
hypochlorous acid
zinc nitrate
silver phosphate
ammonium carbonate
lithium bromide
sulfur dioxide
sodium sulfite
magnesium hydrogen sulfate