Chem. 152
Dr. Saidane
Stoichiometry Key
1. How many moles of oxygen are needed to react with 0.38 mol of C3H8 according to the
following equation? C3H8 + 5O2  3 CO2 + 4H2O
0.38 mol C3H8 x(5 mol O2/1 mol C3H8) = 1.9 mol O2.
2. How many moles of hydrogen gas will be produced if 0.44 mol of CaH2 reacts according to the
following equation? CaH2 + 2 H2O  Ca(OH)2 + 2 H2
0.44 mol CaH2 x (2 mol H2/ 1 mol CaH2) = 0.88 mol H2.
3. What mass of carbon monoxide must react with oxygen to produce 0.69 g of carbon dioxide? 2
CO + O2  2 CO2
0.69 g CO2x(1 mol CO2/44 g CO2)x( 2 mol CO/ 2 mol CO2)x(28 g CO/1 mol CO)=0.44 g CO
4. How many grams of ozone (O3) must decompose to produce 0.87 g of oxygen?
2 O3  3 O2
0.87 g O2x(1mol O2/32 g O2)x(2 mol O3/3 mol O2)x(48g O3/1mol O3)= 0.87 g O3.
5. Find the mass of benzene (C6H6) required to produce 2.66 L of carbon dioxide gas at STP from
the reaction 2 C6H6 + 15 O2  6 H2O + 12 CO2
2.66 L CO2 x (1mol CO2/ 22.4 L CO2) x (2 mol C6H6/ 12 mol CO2) x (78 g C6H6/ 1 mol C6H6) =
1.54 g C6H6
6. What volumes of sulfur dioxide and dihydrogen sulfide gases are necessary to produce 11.4 L
of water vapor? The balanced equation is SO2 + 2 H2S  3 S + 2 H2O.
11.4 L H2O x (1 L SO2/ 2 L H2O) = 5.7 L SO2
11.4 L H2O x (2 L SO2/ 2 L H2O) = 11.4 L H2O
7. Nitroglycerin decomposes explosively to produce carbon dioxide water, nitrogen, and oxygen.
What volumes of nitrogen and oxygen are produced is 4.3 L of carbon dioxide is produced?
The balanced equation is: 4 C3H5(NO3)3  12 CO2 + 10 H2O + O2 + 6 N2
4.3 L CO2 x ( 6 L N2/12 L CO2) = 2.15 L N2
4.3 L CO2x ( 1 L O2/12 L CO2) = 0.36 L O2
8. In a spacecraft, the carbon dioxide exhaled by astronauts can be removed by its reaction with
lithium hydroxide, LiOH, according to the following reaction:
CO2 (g) + 2 LiOH (s)
 Li2CO3 (s) + H2O (l)
How many moles of lithium hydroxide are required to react with 20 mol of CO2, the average
amount exhaled by a person each day?
20 mol CO2 x (2 mol LiOH/1 mol CO2)= 40 mol LiOH
9. In photosynthesis, plants use energy from the sun to produce glucose, C 6H12O6, and oxygen
from the reaction of carbon dioxide and water. What mass, in grams, of glucose is produced
when 3.00 mol of water react with carbon dioxide?
6 CO2(g) + 6 H2O(l)  C6H12O6(s) + 6 O2(g)
3.00 mol H2Ox(1 mol C6H12O6/ 6 mol H2O)x (180 g C6H12O6/ 1 mol C6H12O6)= 90 g C6H12O6
10. The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.
The reaction is run using 824 g of ammonia.
NH3 (g) + O2 (g)  NO (g) +
H2O (g)
a) How many moles of water and nitrogen monoxide are formed?
824 g NH3 x (1 mol NH3/17 g NH3)x(1 mol H2O/1 mol NH3)= 48.5 mol H2O
824 g NH3 x (1 mol NH3/17 g NH3)x(1 mol NO/1 mol NH3)= 48.5 mol NO
b) How many grams of water and nitrogen monoxide are formed?
48.5 mol H2O x (18 g H2O/1 mol H2O) = 872.5 g H2O
48.5 mol NO x (30 g NO/1 mol NO) = 1455 g NO
11. Identify the limiting reactant when 5.87g of Mg(OH)2 reacts with 12.84g of HCl to form
MgCl2 and water. Mg(OH)2 + 2 HCl  MgCl2 + H2O
5.87 g Mg(OH)2x(1mol Mg(OH)2/58.31 g Mg(OH)2x (2 mol HCl/ 1 mol Mg(OH)2)x
(36.45 g HCl/1 mol HCl)= 7.34 g HCl needed < 12.84 g HCl given, Mg(OH)2 is limiting.
12. Identify the limiting reactant when 6.33g of H2SO4 reacts with 5.92g of NaOH to produce
Na2SO4 and water. H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
5.92 g NaOH x (1mol NaOH/40 g NaOH)x(1 mol H2SO4/2molNaOH)x(98 g H2SO4/1molH2SO4) =
7.25 g H2SO4 needed > 6.33 g H2SO4 given, H2SO4 is limiting.
13. If 4.1 g of Cr is heated with 9.3g of Cl2, what mass CrCl3 will be produced?
2 Cr + 3 Cl2  2 CrCl3
4.1 g Cr x ( 1mol Cr/52 g Cr) x (3 mol Cl2/2 mol Cr)x(70.9 g Cl2/1mol Cl2)= 8.35 g Cl2 needed.
Cl2 needed < Cl2 given (9.3 g ) , Cr is limiting.
4.1 g Cr x (1mol Cr/52 g Cr)x(2 mol CrCl3/2 mol Cr)x(158.35 g CrCl3/1molCrCl3)= 12.5g CrCl3.
14. What mass of SO3 is produced from the reaction of 12.4g of SO2 and 3.45g of O2?
2 SO2 + O2  2 SO3
12.4 g SO2x(1mol SO2/64 g SO2)x(1 mol O2/2 mol SO2)x(32 g O2/1mol O2)=3.1 g O2 needed
O2 needed < O2 Given, SO2 limiting.
12.4 g SO2 x (1mol SO2/64 g SO2) x(2 mol SO3/2mol SO2)x(80 g SO3/1 mol SO3)= 15.5 g SO3.
15. If 21.4g of aluminum is reacted with 91.3g of Fe2O3, the products will be Al2O3 and iron.
What mass of iron will be produced?
2 Al + Fe2O3  Al2O3 + 2 Fe
21.4 g Alx(1mol Al/27 g Al)x(1 mol Fe2O3/2mol Al)x( 160 g Fe2O3/ 1 mol Fe2O3) = 63.4 g Fe2O3 needed.
Fe2O3 needed < Fe2O3 given, Al is limiting
21.4 g Al x (1 mol Al/27 g Al)x(1 mol Al2O3/2 mol Al)x(102 g Al2O3/1 mol Al2O3) = 40.4 g Al2O3
16. Calculate the percent yield for the reaction between 6.92g of K and 4.28g of O2 if 7.36g of
4 K + O2  2 K2O
K2O is produced.
6.92 g K x (1mol K/39 g K)x(1 mol O2/4mol K)x (32 g O2/1 mol O2)= 1.42 g O2 needed.
O2 needed < O2 given. K is limiting.
6.92 g K(1 mol K/39 g K)x(2 mol K2O/4 mol K)x(94 g K2O/1 mol K2O)= 8.34 g K2O=Theoretical.
% yield = (actual/theoretical)x100= (7.36 g /8.34 g)x100=88.3%
17. Determine the percent yield for the reaction between 28.1g of Sb4O6 and excess C if 17.3g of
Sb is recovered with an unknown amount of CO.
Sb4O6 + 6 C  4 Sb + 6 CO
28.1 g Sb4O6 x (1mol Sb4O6/583.2 g Sb4O6)x(4 mol Sb/1 mol Sb4O6)x(121.8 g Sb/1 mol Sb)= 23.5 g
Sb = theoretical yield.
% yield = (actual/theoretical) x 100= (17.3/23.5)x100= 73.7 %.
18. Some rocket engines use a mixture of hydrazine, N2H4, and hydrogen peroxide, H2O2, as the
propellant.
a)
N2H4 (l) + 2 H2O2 (l)  N2 (g) + 4 H2O (g)
Which is the limiting reactant in this reaction when 0.750 mol of N2H4 is mixed with
0.500 mol of H2O2?
0.750 mol N2H4 x (2 mol H2O2/1 mol N2H4)= 1.5 mol H2O2 needed.
Needed> given. H2O2 limiting
b)
How much of the excess reactant, in moles, remains unchanged?
0.5 mol H2O2 x (1 mol N2H4/ 2 mol H2O2)= 0.25 mol N2H4 used.
0.750 -0.25 = 0.50 mol N2H4 in excess remains unchanged.
c)
How much of each product, in grams, is formed?
0.5 mol H2O2 x (1 mol N2/2 mol H2O2)x(28 g N2/1 mol N2)= 7 g N2
0.5 mol H2O2 x (4 mol H2O/2 mol H2O2)x(18 g H2O/1 mol H2O)= 18 g H2O
19. Methanol can be produced through the reaction of CO and H2 in the presence of a catalyst.
CO (g) + 2 H2 (g) 
CH3OH (l)
If 75.0 g of CO reacts to produce 68.4 g CH3OH, what is the percent yield?
75.0 g CO x(1mol CO/28 g CO)x(1 mol CH3OH/1 mol CO)x (30 g CH3OH/1 mol CH3OH)=80.4 g CH3OH
= theoretical yield.
% yield = (Actual/Theoretical) x 100 = (68.4/80.4)x100= 85.1 %