Answers to The Predicted Yield Questions
1) Answer = 78.7
Explanation
1) Find the limiting reagent (if you forgot how Click Here)
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 Click Here)
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
Explanation
1) Find the limiting reagent (if you forgot how Click Here)
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 Click Here)
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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