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```					Percent Purity, Percent Yield                                         Name: __________________________
Date: _________________

1. The roasting of siderite ore, FeCO3, produces iron (III) oxide:
FeCO3 + O2  Fe2O3 + CO2
a. What is the balanced equation?
4 FeCO3 + O2  2 Fe2O3 + 4 CO2
b. A 15.0 g FeCO3 sample is 42.0% pure. What mass of Fe2O3 can the sample produce?
(6.30 g pure FeCO3 will react)
4.34 g Fe2O3
c. A second sample of FeCO3, with a mass of 55.0 g is roasted so as to produce 37.0g of Fe2O3.
What is the percentage purity of FeCO3?
(49.9g pure FeCO3)
49.9g / 55.0 g * 100% = 97.6 %
d. A 35.0 g sample of pure FeCO3 produces 22.5 g of Fe2O3. What is the percentage yield of the
reaction?
(24.1g pure Fe2O3 produced)
22.5g / 24.1 g *100% = 93.3%
e. What mass of siderite ore with a purity of 62.8% is needed to make 1.00 kg of Fe2O3?
2310 g

2. A 100. g sample of impure FeS2 is roasted to produce Fe2O3 + SO2:
4 FeS2 + 11 O2  2 Fe2O3 + 8 SO2
If 4.50 L of SO2 is collected at STP, what percentage of FeS2 is in the sample?
(find the mass of FeS2 required to make 4.50L of SO2 = 12.1 g FeS2)
12.1%

3. ***This question is quite challenging***
The reaction:
SiO2 (g) + HF (g)  SiF4 (g) + 2 H2O (g)
produces 2.50g of H2O when 12.20g of SiO2 is treated with a small excess of HF?
a. What is the balanced equation?
SiO2 + 4 HF  SiF4 + 2 H2O
b. What mass of SiF4 is formed?
7.23g
c. What mass of SiO2 is left unreacted?
8.03g
d. What is the percentage yield of SiF4?
34.2%

4. When 5.000 kg of malachite ore containing 4.30% of malachite, Cu2(OH)2CO3, is heated, the product
is copper (II) oxide:
Cu2(OH)2CO3  CO2 + CuO + H2O
a. What is the balanced equation?
Cu(OH)2CO3  CO2 + 2 CuO + H2O
b. If the reaction has an 84.0% yield, how many grams of CuO are produced?
1.30 x 102 g
c. If the decomposition reaction has an 87.0% yield, what mass of ore containing 3.70%
malachite is required to produce 100.0g of CuO?
4.32 x 103 g

5. A mine produces a silver ore named argentite, Ag2S. The ore is smelted according to the overall
reaction
Ag2S + C + O2  Ag + CO2 + SO2
a. What is the balanced equation?
Ag2S + C + 2 O2  2 Ag + CO2 + SO2
b. A 250.0 kg load of argentite ore contains 0.135% pure Ag2S. What mass of silver metal can
be produced from the load of core?
294 g
c. A 76.4 g test sample of ore from a new ore vein produces 0.261g of pure silver. What is the
percentage of pure argentite in the ore?
0.392%
d. A sample of pure Ag2S has a mass of 152.6 g. When smelted, the sample produces 117.4 g of
pure Ag. What is the percentage yield of the smelting process?
88.38%
e. What mass of ore containing 0.795 % Ag2S is required to produce a 50.0 kg ingot of silver
metal?
7.22 x 106 g
f. If 89.2% of the Ag2S present is extracted from 3.50x104 kg of ore containing 1.86% Ag2S,
what mass of silver metal can be produced?
5.05 x 105 g

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