## Answers to The Predicted Yield Questions

1)

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) =

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)

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)

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)

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)

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)

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)

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)

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)

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

**Answer = 78.7**

Explanation1) 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**

Explanation1) 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**

Explanation1) 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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