Stoich and % Yield Problems
1.
How many grams of chlorine gas are needed in the production of 200.0 grams of
bromine gas from sodium bromide?
Cl2 + 2NaBr ® 2NaCl + Br2
200.0
=1.2515 mol Br2 = 1.2515 mol Cl2
159.808
1.2515 (70.906) = 88.74 g Cl2
2.
Calculate the percent yield of a reaction that consumes 900.0 grams of
potassium chlorate in the preparation of 90.0 grams of oxygen?
2KClO3 ® 2KClO2 + O2
1O2
900.0
= 7.344 mol KClO3
= 3.672 mol O2
2KClO3
122.548
3.672 (31.998) = 117.49 g of O2 (TY)
90.0
(100) = 76.6% yield
117.49
3.
In the reaction of 84.0 g of copper (II) oxide with hydrogen gas in a reaction
with a 92.4% yield, Determine the mass of copper produced?
CuO + H 2 ® Cu + H 2O
84.0
= 1.056 mol CuO = 1.056 mol Cu (TY)
79.545
1.056 (63.546) = 67.1 g Cu (TY)
AY
(100) = 92.4 %
67.4
4.
AY = 62.0 g of Cu
How many grams of sodium hydroxide will react with 1.50 x 10-3 g of phosphoric
acid?
3NaOH + H 3PO4 ® 3H 2O + Na3PO4
3NaOH
0.00150
= 1.53 x 10-5 mol H3PO4
= 4.59 x 10-5 mol NaOH
1H 3PO4
97.9937
4.59 x 10-5 (39.9969) = 1.84 x 10-3 g of NaOH
5.
If 320.0 grams of sodium carbonate react with calcium hydroxide in a 74.67%
efficient reaction, how many grams of sodium hydroxide are formed?
Na2CO3 + Ca(OH )2 ® 2NaOH + CaCO3
2NaOH
320.0
= 3.019 mol Na2CO3
= 6.038 mol NaOH
1Na2CO3
105.988
6.038 (39.9969) = 241.5 g NaOH (TY)
AY
AY = 180.3 g of NaOH
(100) = 74.67%
241.5
6.
How many grams of sodium iodide are produced by the decomposition of 60.0 g
of sodium iodate if the reaction is 89.4% effective?
_ 2 _ NaIO3 ® _ 2 _ NaI + _ 3_O2
60.0
= 0.3032 mol NaIO3 = 0.3032 mol NaI
197.887
0.3032 (149.89) = 45.4466 g NaI (TY)
AY
AY = 40.6 g NaI
(100) = 89.4%
45.4466
7.
If 10.0 g of aluminum sulfide are produced by the 94.2% effective reaction of
aluminum and sulfur, how many grams of sulfur were needed?
_ 2 _ Al + _ 3_ S ® ___ Al2 S3
Don’t worry about this guy if you can’t figure it out
10.0
TY = 10.6 g Al2S3
(100) = 94.2%
TY
10.6
= 0.07059 mol Al2S3
150.159
3S
0.07059
0.211775 mol S
Al2 S3
0.211775 (32.065) = 6.79 g of S
8.
If 2.50 g of cupric sulfide is produced by decomposing cupric sulfate upon
heating, how many grams of cupric sulfate was necessary if the reaction is
90.00% efficient?
___CuSO4 ® ___ CuS + _ 2 _O2
Don’t worry about this guy if you can’t figure it out
2.50
TY = 2.7777 g of CuS
(100) = 90.00%
TY
2.7777
= 0.0290529 mol CuS = 0.0290529 mol CuSO4
95.611
0.0290529 (159.607) = 4.64 g of CuSO4
9. 125.0 g of FeS react to form 167.3 g of FeCl2:
a. Determine the % yield of the reaction.
b. Calculate, based on that percent yield, the moles of H2S that will be
produced.
FeS(s) + 2HCl(aq) ® FeCl2(aq) + H 2S(g)
125.0
= 1.4219 mol FeS = 1.4219 mol FeCl2
87.910
1.4219 (126.751) = 180.227 g FeCl2 (TY)
167.3
(100) = 92.83% yield
180.227
10. If 1.487 moles of lithium oxide react with water through a 79.4% effective reaction,
determine the mass of lithium hydroxide produced.
Li2O(s) + H 2O(l) ® 2LiOH(aq)
2LiOH
= 2.974 mol LiOH
1Li2O
1.487
2.974 (23.9479) = 71.22 g LiOH (TY)
AY
(100) = 79.4 %
71.22
AY = 56.5 g LiOH
11. In order to produce 100.0 g of NaCl based on the following 84.3 % efficient reaction,
determine the moles of each reactant required.
Na 2SO4(aq) + CaCl2(aq) ® CaSO4(s) + 2NaCl(aq)
100.0
= 1.711 mol NaCl (AY)
58.443
1.711
(100) = 84.3%
TY
2.0297
2.0297
TY = 2.0297 mol NaCl
1CaCl2
= 1.01 mol CaCl2
2NaCl
1Na2 SO4
= 1.01 mol Na2SO4
2NaCl
12. 0.941 moles of calcium hydroxide react to form 30.55 g of water. Determine the
percent yield of the reaction.
Ca(OH)2(aq) + 2HCl(aq) ® CaCl2(aq) + 2H 2O(l)
0.941
2H 2O
= 1.882 mol H2O
1Ca(OH )2
1.882 (18.0148) = 33.9038 g H2O (TY)
30.55
(100) = 90.11 % yield
33.9038
13. What mass of ethane must react in order to produce 4.73 moles of carbon dioxide if
the following reaction has a 91.8 % yield?
2C2H 6 + 7O 2 ® 4CO 2 + 6H 2O
4.73
(100) = 91.8 %
TY
5.1525
TY = 5.1525 mol CO2
2C2 H 6
= 2.57625 mol C2H6
4CO2
2.57625 (30.0694) = 77.5 g of C2H6
14. 4.84 moles of oxygen react in the following 87.9 % yield reaction. Calculate the
mass of each product produced.
2Cu2S + 3O 2 ® 2Cu2O + 2SO2
4.84
2(Either)
= 3.22666 mol Cu2O and SO2 (TY)
3O2
AY
(100) = 87.9 %
3.22666
AY = 2.83624 mol of Cu2O and SO2
2.83624 (143.091) = 406 g of Cu2O
2.83624 (64.063) = 182 g of SO2