If you got less than 5/5 correct refer to the links below.
If your errors were due to incorrectly calculating the Molar Masses, go to "Practice with Molar Masses."
If your errors were due to finding the incorrect Limiting Reagent, go to "How to Find the Limiting Reagent."
If you errors were due to incorrectly applying the Formula for Using Limiting Reagents, go to "Using Limiting Reagents."

ANSWERS:

1) For the balanced equation shown below, if 64.7 grams of C2H4O were reacted with 59.2 grams of O2, how many grams of CO2 would be produced? 65.12

2C2H4O+5O2=>4CO2+4H2O

Once you have found the limiting reagent = 02 (whose moles = 1.85) then you simply apply the formula.
Theoretical yield in moles = (moles of the limiting reagent) * (coefficient of the product/coefficient of the limiting reagent)

Theoretical yield in moles = (1.85) * (4/5) = 1.48

Theoretical yield in grams = (theoretical yield in moles) * (the molar mass of the product)

Theoretical yield in grams = (1.48) * (44) = 65.12

2) For the balanced equation shown below, if 91.2 grams of C6H5Cl were reacted with 66.7 grams of O2, how many grams of H2O would be produced? 18.72

C6H5Cl+4O2=>6CO+2H2O+HCl

Once you have found the limiting reagent = 02 (whose moles = 2.08) then you simply apply the formula.
Theoretical yield in moles = (moles of the limiting reagent) * (coefficient of the product/coefficient of the limiting reagent)

Theoretical yield in moles = (2.08) * (2/4) = 1.04

Theoretical yield in grams = (theoretical yield in moles) * (the molar mass of the product)

Theoretical yield in grams = (1.04) * (18) = 18.72

3) For the balanced equation shown below, if 39.9 grams of CH3Cl were reacted with 85.4 grams of O2, how many grams of Cl2 would be produced? 28.045

4CH3Cl+7O2=>4CO2+6H2O+2Cl2

Once you have found the limiting reagent = CH3Cl (whose moles = .790) then you simply apply the formula.

Theoretical yield in moles = (moles of the limiting reagent) * (coefficient of the product/coefficient of the limiting reagent)

Theoretical yield in moles = (.790) * (2/4) = .395

Theoretical yield in grams = (theoretical yield in moles) * (the molar mass of the product)

Theoretical yield in grams = (.395) * (71) = 28.045

4) For the balanced equation shown below, if 31.0 grams of PCl5 were reacted with 16.2 grams of H2O, how many grams of H3PO4 would be produced? 14.504

PCl5+4H2O=>H3PO4+5HCl

Once you have found the limiting reagent = PCl5 (whose moles = .148) then you simply apply the formula.

Theoretical yield in moles = (moles of the limiting reagent) * (coefficient of the product/coefficient of the limiting reagent)

Theoretical yield in moles = (.148) * (1/1) = .148

Theoretical yield in grams = (theoretical yield in moles) * (the molar mass of the product)

Theoretical yield in grams = (.148) * (98) = 14.504

5) For the balanced equation shown below, if 38.4 grams of C6H12O2 were reacted with 165 grams of O2, how many grams of H2O would be produced? 35.64

C6H12O2+8O2=>6CO2+6H2O

Once you have found the limiting reagent = C6H12O2 (whose moles = .33) then you simply apply the formula.

Theoretical yield in moles = (moles of the limiting reagent) * (coefficient of the product/coefficient of the limiting reagent)

Theoretical yield in moles = (.33) * (6/1) = 1.98

Theoretical yield in grams = (theoretical yield in moles) * (the molar mass of the product)

Theoretical yield in grams = (1.98) * (18) = 35.64