## Answers to The Predicted Yield Questions

1) Answer = 78.7
Explanation
1) Find the limiting reagent (if you forgot how
If you did find the limiting reagent you would know that it is
C2H6S with a molarity of apox .895
2) Multiply by the molarity by the ratio between the limiting reagent and the desired product
1:2
2(.895) = 1.79
3) Find the GFW of the product desired (If you forgot how
GFW of CO2 = 44
4) Multiply the product of the ratio x limiting reagent by the GFW of the product desired
44(1.79) =
78.7

2) Answer = 4.4
Explanation
1) Limting reagent
O2 w/ molarity = .1
2) Multiply by ratio
1(.1) = .1
3) Find GFW of product desired
GFW of CO2 = 44
3) Multiply by GFW of product desired
.1(44) = 4.4
Predicted Yield = 4.4 grams of CO2

3) Answer = 35.7
Explanation
1) Limiting Reagent
O2 w/ molarity = .94
2) Multiply by ratio
7:6 or 1:6/7
6/7(.94) = .81
3) Find GFW of product desired
GFW of CO2 = 44
4) multiply by GFW of product desired
44(.81) = 35.64, round up 35.7

4) Answer = 29 or 28.9
Explanation
1) Limiting Reagent
H2O w/ molarity = .86
2) Multiply by ratio
1/3(.86) = .29
3) Find GFW of desired product
GFW of CaCO3 = 100
4) Multiply by GFW of product desired
.29(100) = 29
Predicted Yield is 29g of CaCO3

5) Answer = 81.46
1) Limiting Reagent
Fe w/ Molarity = .214
2) Multiply by ratio
3(.214) = .642
3) Find GFW of desired product
GFW of FeCl2 = 126.9
4) Multiply by GFW of product desired
126.9(.642) = 81.46
Predicted Yield is 81.46g of FeCl2

6) Answer = 3.7
Explanation
1) Limiting Reagent
O2 w/ Molarity = .24
2) Multiply by ratio
6/7(.214) = .205
3) Find GFW of desired product
GFW of H2O = 18
4) Multiply by GFW of product desired
18(.2) = 3.69
Predicted Yield is 3.7g of H2O

7) Answer = 6.4
Explanation
1) Limiting Reagent
S w/ Molarity = .04
2) Multiply by ratio
1(.04) = .04
3) Find GFW of desired product
GFW of Cu2S = 160
4) Multiply by GFW of product desired
160(.04) = 81.46
Predicted Yield is 6.4g of FeCl2

8) Answer = 129
Explanation
1) Limiting Reagent
Al w/ Molarity = .966
2) Multiply by ratio
1(.966) = .966
3) Find GFW of desired product
GFW of FeCl2 = 133.35
4) Multiply by GFW of product desired
133.35(.642) = 128.8
Predicted Yield is 128.8g of AlCl3

9) Answer = 220.5
Explanation
1) Limiting Reagent
Al w/ Molarity = 2.94
2) Multiply by ratio
1/2(2.94) = 1.47
3) Find GFW of desired product
GFW of Al2S3 = 150
4) Multiply by GFW of product desired
150(2.94) = 220.5
Predicted Yield is 220.5g of AlCl3

10) Answer = 48.3
Explanation
1) Limiting Reagent
FeCl3 w/ Molarity = .25
2) Multiply by ratio
3/2(.25) = .38
3) Find GFW of desired product
GFW of FeCl2 = 126.9
4) Multiply by GFW of product desired
126.9(.38) = 48.2
Predicted Yield is 48.2g of FeCl2
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