# Percent Yield - Chemistry

Document Sample

```					                                     Percent Yield - Chemistry
Name: ____________________________________

Percent Yield = (actual yield/theoretical yield) x 100

1. A student adds 200.0g of C7H6O3 to an excess of C4H6O3, this produces C9H8O4 and C2H4O2. Calculate the
percent yield if 231 g of aspirin (C9H8O4) is produced.

C7H6O3 + C4H6O3  C9H8O4 + C2H4O2

≈90%

2. According to the following equation calculate the percentage yield if 550.0 g of toluene (C7H8) added to an
excess of nitric acid (HNO3) provides 305 g of the p-nitrotoluene product.

C7H8 + HNO3  C7H7NO2 + H2O

37.2%

3. Quicklime, CaO, can be prepared by roasting limestone, CaCO3, according to the chemical equation below.
When 2.00 x 103 g of CaCO3 are heated, the actual yield of CaO is 1.50 x 103 g. What is the percentage yield?

CaCO3  CaO + CO2

93.8%
4. Aluminum reacts with an aqueous solution containing excess copper (II) sulfate. If 1.85 g Al reacts and the
percentage yield of Cu is 56.6%, what mass of Cu is produced?

Al + CuSO4         Cu + Al2(SO4)3

3.70g

5. The combustion of methane produces carbon dioxide and water. Assume that 2.0 mol of CH4 burned in the
presence of excess air. What is the percentage yield if the reaction produces 87.0 g of CO2?

99%

6. How many atoms of chromium are in 23.2g of Cr ore that is 73.4% pure?
7. What is the percent yield of this reaction if 14 moles of ZnO are produced from the reaction of 256 grams of
O2 with excess Zn?
Zn + O2 → ZnO

8. If the percent yield is 40%, how many atoms of O are produced from 6.02×1023 formula units of KClO3(s)?
KClO3(s) → KCl(s) + O2(g)

9. At STP, 9.83×1024 formula units of KClO3(s) decompose into KCl and 82 L of O2(g). What is the percent
yield?

KClO3(s) → KCl(s) + O2(g)
10. What is the percent yield of the following reaction if 60 grams of CaCO3 is heated to give 15 grams of
CaO?

CaCO3→ CaO + CO2
Percent Yield - Chemistry
Name: ____________________________________

Percent Yield = (actual yield/theoretical yield) x 100

1. A student adds 200.0g of C7H6O3 to an excess of C4H6O3, this produces C9H8O4 and C2H4O2. Calculate
the percent yield if 231 g of aspirin (C9H8O4) is produced.

C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2

1 C7H6O3 + 1 C4H6O3 → 1 C9H8O4 + 1 C2H4O2

200.0 g C7H6O3     1 mol C7H6O3          1 mol C9H8O4      180.1582 g C9H8O4          = 260.869 g C9H8O4
138.1214 g C7H6O3      1 mol C7H6O3      1 mol C9H8O4

231 g C9H8O4          x 100 =      0.8855 %
260.869 g C9H8O4

2. According to the following equation calculate the percentage yield if 550.0 g of toluene (C7H8) added to
an excess of nitric acid (HNO3) provides 305 g of the p-nitrotoluene (C7H7NO2) product.

C7H8 + HNO3 → C7H7NO2 + H2O

550.0 g C7H8      1 mol C7H8    1 mol C7H7NO2 137.1373 g C7H7NO2 = 818.595 g C7H7NO2
92.1402 g (C7H8) 1 mol C7H8        1 mol C7H7NO2

305 g C7H7NO2   =       37.259 %
818.595 g C7H7NO2)
3. Quicklime, CaO, can be prepared by roasting limestone, CaCO3, according to the chemical equation
below. When 2.00 x 103 g of CaCO3 are heated, the actual yield of CaO is 1.50 x 103 g. What is the
percentage yield?

CaCO3 → CaO + CO2                      1 CaCO3  1 CaO + 1 CO2

2000 g CaCO3       1 mol CaCO3         1 mol CaO            56.079 g CaO = 1120.594 g CaO
100.088 g CaCO3     1 mol CaCO3          1 mol CaO

Ca 1 x 40.08 = 40.08
C 1 x 12.011 = 12.011
O 3 x 15.999 = 47.997
100.088 g CaCO3
Ca 1 x 40.08 = 40.08
O 1 x 15.999 = 15.999
56.079 g CaO

1500 x 100 =         133.858 %
1120.594

4. Aluminum reacts with an aqueous solution containing excess copper (II) sulfate. If 1.85 g Al reacts and
the percentage yield of Cu is 56.6%, what mass of Cu is produced?

2 Al + 3 CuSO4 → 3 Cu           + Al2(SO4)3

1.85 g Al    1 mol Al      3 mol Cu        63.546 g Cu    56.6   = 3.699 g Cu
26.982 g Al    2 mol Al         1 mol Cu      100
5. The combustion of methane produces carbon dioxide and water. Assume that 2.0 mol of CH4 burned in
the presence of excess air. What is the percentage yield if the reaction produces 87.0 g of CO2?

CH4 + 2 O2 → CO2 + 2 H2O

2.0 mol CH4     1 mol CO2   44.009 g CO2    = 88.018 g CO2
1 mol CH4   1 mol CO2

87.0 g of CO2 x 100 = 0.988%
88.018 g CO2

6. How many atoms of chromium are in 23.2 g of Cr ore that is 73.4 % pure?

23.2 g Cr     73.4    1 mol Cr     6.02 x 1023 atoms   = 1.972 x 1023 atoms Cr
100    51.996 g Cr       1 mol Cr
7. What is the percent yield of this reaction if 14 moles of ZnO are produced from the reaction of 256 grams
of O2 with excess Zn?
2 Zn + O2 → 2 ZnO

256 g O2      1 mol O2       2 mol ZnO     81.389 g ZnO     = 1302.305 g ZnO
31.998 g O2      1 mol O2       1 mol ZnO

14 moles of ZnO      1 mol ZnO       =   0.172 mol ZnO
81.389 g ZnO

0.172 mol ZnO     = 0.0132%
1302.305 g ZnO

8. If the percent yield is 40%, how many atoms of O are produced from 6.02×1023 molecules of KClO3(s)?

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

6.02 ×1023 molecules of KClO3         1 mol KClO3          3 mol O2   6.02 ×1023 molecules O2 =
6.02×1023 molecules KClO3 1 mol KClO3     1 mol O2

= 1.806 ×1024 molecules O2     40    = 7.224 ×1023 molecules O2
100

9. At STP, 9.83×1024 formula units of KClO3(s) decompose into KCl and 82 L of O2(g). What is the percent
yield?

KClO3(s) → KCl(s) + O2(g)
10. What is the percent yield of the following reaction if 60 grams of CaCO3 is heated to give 15 grams of
CaO?

CaCO3→ CaO + CO2

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 296 posted: 4/18/2012 language: English pages: 9