Created by Laura Peck, 2011
Modern Chemistry
STOICHIOMETRIC REVIEW
Example/Reminder
 Ex.1) Given the chemical equation:
Na2CO3(aq) + Ca(OH)2(s)
2NaOH(aq) + CaCO3(s)
If we start with 2g Na2CO3 – how many grams NaOH
are produced?
Grams
start
2g Na2CO3
1 mol
Molar mass
(start)
1 mol
105.967 g Na2CO3
(ratio)
Mol final
Mol start
2 mol
1 m0l
Grams final
(molar
mass)
39.987 g NaOH
1 mol
2 x 1 x 2 x 39.987 / (105.965 x 1 x 1) = 1.509 g NaOH
Practice problems
 1) Hydrogen and oxygen react under a specific
set of conditions to produce water according to
the following: 2H2(g) + O2(g)
2H2O(g)
 A) How many grams of hydrogen would be
required to produce 5.0 g of water?
5.0 g H2O
1 mol H2O
2 mol
2.014 g H2
= 0.559 g H2
18.01 g H2O 2 mol
 B) How many grams of oxygen would be required
to produce 5.0 g of water?
5.0 g H2O
1 mol H2O 1 mol O2
31.98 g O2
18.01 g H2O 2 mol H2O 1 mol O2
= 4.44 g O2
 2) If 4.5 g of ethane, C2H6, undergo combustion
according to: 2C2H6 + 5O2
4CO2 + 6H2O
 A) how many grams of oxygen are required?
4.5 g C2H6
1 mol C2H6
5 mol O2 31.98 g O2
30.062 g C2H6 2 mol C2H6 1 mol O2
= 11.967 g O2
 B) how many grams of carbon dioxide are produced?
4.5 g C2H6 x 1 mol C2H6 x 4 mol CO2 x 43.99g CO2
30.062 g C2H6 x 2mol C2H6 x 1 mol CO2
= 13.17g CO2
 C) how many grams of water are produced?
4.5g C2H6 x
1 mol C2H6 x 6 mol H2O x 18.01g H2O
30.062g C2H6 x 2mol C2H6 x 1 mol H2O
= 8.08g H2O
 3) Sodium chloride is produced from its element through
a synthesis reaction. 2Na(s) + Cl2(g)
2NaCl(s)
 A) What mass of each reactant would be required
to produce 3.4g of sodium chloride?
3.4g NaCl x 1 mol NaCl x 2 mol Na x 22.989g Na
58.439g NaCl x 2mol NaCl x 1 mol Na
3.4g NaCl x 1 mol NaCl x 1 mol Cl2 x 70.9g Cl2
58.439g NaCl x 2mol NaCl x 1 mol Cl2
= 1.34 g Na
= 2.06g Cl
 B) What mass of each reactant would be required
to produce 13.4g of sodium chloride?
13.4 g NaCl x 1 mol NaCl
58.43 g NaCl
13.4g NaCl x 1 mol NaCl
58.43 g NaCl
x 2 mol Na x 22.98g Na
x 2mol NaCl x 1 mol Na
x 1 mol Cl2 x 70.9 Cl2
x 2 mol NaCl x 1 mol Cl2
= 5.27g Na
= 8.1 g Cl2
 4) Iron is generally produced from iron ore
through the following reaction in a blast furnace:
2Fe2O3(s) + 6CO(g)
4Fe(s) + 6CO2(g)
 A) If 4.0 kg of Fe2O3 are available to react how many
grams of CO are needed?
4000 g Fe2O3 x 1 mol Fe2O3 x 6 mol CO x 28.009 g CO
159.687 g Fe2O3 x 2mol Fe2O3 x 1 mol CO
= 2104.8 g CO
 B) If 4.0 kg of Fe are produced, how many grams of Fe2O3
are needed?
4000 g Fe x 1 mol Fe x 2 mol Fe2O3 x 159.687 g Fe2O3
55.845 g Fe x 4 mol Fe
x 1 mol Fe2O3
= 5718.93 g Fe2O3
 5) Methanol, CH3OH, is an important industrial
compound that is produced from the following
reaction: CO(g) + 2H2(g)
CH3OH(g)
 A) What mass of each reactant would be needed to
produce 100.0 kg of methanol?
100000 g CH3OH x 1 mol CH3OH x 1 mol CO x 28.0 g CO
32.037 g CH3OH x 1 mol CH3OH x 1 mol CO
100000 g CH3OH x 1 mol CH3OH x 2 mol H2 x 2.014 g H2
32.037 g CH3OH x 1 mol CH3OH x 1mol H2
= 87,398.94 g CO
= 12,572.96 g H2
 B) How many mols of H2 were consumed to generate
100.0 kg of methanol?
12572.96 g H2
x 1 mol H2
2.014 g H2
= 6242.78 mol H2
 5) Nitrogen combines with oxygen in the atmosphere
during lightning flashes to form nitrogen monoxide, NO,
which then reacts further with O2 to produce nitrogen
dioxide, NO2 N2 + O2 2NO
2NO + O2 2NO2
 A) What mass of NO2 is formed when NO reacts with 384
g of O2?
384 g O2 x 1 mol O2 x 2 mol NO2 x 46.065 g NO2
31.998 g O2 x 1 mol O2 x 1 mol NO2
= 1105.63 g NO2
 B) how many grams of NO are required to react with 455g
of O2?
455 g O2 x 1 mol O2 x 2 mol NO x 30.066 g NO
31.989 g O2 x 1 mol O2 x 1 mol NO
= 855.29 g NO
 6) As early as 1938, the use of NaOH was suggested as a
means of removing CO2 from the cabin of a spacecraft
according to the following reaction:
2NaOH + CO2
Na2CO3 + H2O
 A) If the average human body discharges 925.0 g of CO2
per day, how many grams of NaOH are needed each day
for each person in the spacecraft?
925.0 g CO2 x 1 mol CO2 x 2 mol NaOH x 39.995 g NaOH
44.008 g CO2 x 1 mol CO2 x 1 mol NaOH
= 1681.3 g NaOH
 B) How many mols of NaOH were needed for #A?
1681.3 g NaOH x 1 mol NaOH
39.995 g NaOH
= 42.03 mol NaOH
 7) The double-replacement reaction between silver
nitrate and sodium bromide produces silver bromide, a
component of photographic film.
AgNO3 + NaBr
AgBr + NaNO3
 A) If 4.5 g of silver nitrate reacts, what mass of sodium
bromide is required?
4.5 g AgNO3 x 1 mol AgNO3 x 1 mol NaBr x 102.88 g NaBr
169.932 g AgNO3 x 1mol AgNO3 x 1 mol NaBr
= 2.72 g NaBr
 B) If 4.5 mol of silver nitrate reacts, how many mols of
sodium bromide is required?
4.5 mol AgNO3 x 1 mol NaBr
1 mol AgNO3
or
1 mol NaBr = x
1mol AgNo3 4.5
= 4.5 mol NaBr
 8) In a soda-acid fire extinguisher, concentrated sulfuric
acid reacts with sodium hydrogen carbonate to produce
carbon dioxide, sodium sulfate and water
H2SO4 + 2NaHCO2
2CO2 + Na2SO4 + 2H2O
 A) How many grams of sodium hydrogen carbonate would
be needed to react with 150.0 g of sulfuric acid?
150.0 g H2SO4 x 1 mol H2SO4 x 2 mol NaHCO2 x 68.004 g NaHCO2
98.07 g H2SO4 x 1 mol H2SO4 x 1 mol NaHCO2
= 208.02 g NaHCO2
 B) how many moles of sodium sulfate would be produced
from 150.0 g sulfuric acid?
150.0 g H2SO4 x 1 mol H2SO4 x 1 mol Na2SO4
98.07 g H2SO4 x 1 mol H2SO4
= 1.53 mol Na2SO4
Limiting reagent practice
- find how many grams each reactant can produce
- the one that produces the least is the limiting
 9) Determine the limiting reagent when 2.o mols of
hydrochloric acid react with 2.5 mols of sodium
hydroxide in a neutralization reaction.
HCl + NaOH
NaCl + H2O
2.0 mol HCl x 1 mol NaCl x 58.439 g NaCl
1 mol HCl x 1 mol NaCl
2.5 mol NaOH x 1 mol NaCl x 58.439 g NaCl
1 mol NaOH x 1 mol NaCl
= 116.878 g NaCl
= 146.09 g NaCl
 10) If 2.50 mol of copper and 5.50 mol of silver
nitrate are available to react by single
replacement, identify the limiting reagent.
Cu + AgNO3
Ag + CuNO3
2.5 mol Cu x 1 mol Ag x 107.86 g Ag
1 mol Cu x 1 mol Ag
5.5 mol AgNO3 x 1 mol Ag x 107.86 g Ag
1 mol AgNO3 x 1 mol Ag
= 269.65 g Ag
= 593.23 g Ag
 11) Sulfuric acid reacts with aluminum hydroxide by
double replacement. H2SO4 + Al(OH)2 AlSO4 + 2H2O
 A) If 30.0 g of sulfuric acid react with 25.0 g of aluminum
hydroxide, identify the limiting reagent.
30.0 g H2SO4 x 1 mol H2SO4 x 2 mol H2O x 18.01 g H2O
98.075 g H2SO4 x 1 mol H2SO4 x 1 mol H2O
25.0 g Al(OH)2 x 1 mol Al(OH)2 x 2 mol H2O x 18.01 g H2O
60.992 g Al(OH)2 x 1 mol Al(OH)2 x 1 mol H2O
= 11.02 g H2O
= 14.76 g H2O
 B) Determine the amount in moles of each product formed.
11.02 g H2O x 1 mol H2O
18.01 g H2O
= 0.61 mols H2O
0.61 mols H2O x 1 mol AlSO4
2 mol H2O
or
1 mol AlSO4 = x
2 mol H2O
0.61
= 0.305 mol AlSO4