Answer Key:

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1) For the balanced equation shown below, if the reaction of 16.4 grams of C6H5F produces a 53.6% yield, how many grams of H2O would be produced? 3.2964

C6H5F+4O2=>6CO+2H2O+HF

Theoretical Yield = (36/96) * (16.4) = 6.15 moles
Actual Yield = 329.64/100 = 3.2964 grams
2) For the balanced equation shown below, if the reaction of 69.9 grams of C produces a 84.0% yield, how many grams of Na2S would be produced? 190.827

Na2SO4+2C=>Na2S+2CO2

Theoretical Yield = (78/24) * (69.9) = 227.175 moles
Actual Yield = 1908.27/100 = 190.827 grams

3) For the balanced equation shown below, if the reaction of 77.9 grams of C2H3O2Cl produces a 68.6% yield, how many grams of CO2 would be produced? 49.735

2C2H3O2Cl+3O2=>4CO2+2H2O+2HCl

Theoretical Yield = (176/189) * (77.9) = 72.5 moles
Actual Yield = 4973.5/100 = 49.735 grams
4) For the balanced equation shown below, if the reaction of 1.02 grams of O2 produces a 69.9% yield, how many grams of CO2 would be produced? .713

4C2H3F+11O2=>8CO2+6H2O+2F2

Theoretical Yield = (352/352) * (1.02) = 1.02 moles
Actual Yield = 71.298/100 = .713 grams

5) For the balanced equation shown below, if the reaction of 24.4 grams of C2H3Cl produces a 68.0% yield, how many grams of CO2 would be produced? 23.358

4C2H3Cl+11O2=>8CO2+6H2O+2Cl2

Theoretical Yield = (352/250) * (24.4) = 34.35 moles
Actual Yield = 2335.8/100 = 23.358 grams

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