75.4 grams of C2H3Br3 reacts with 47.1 grams of O2 to produce CO2. 4C2H3Br3 + 11O2 → 8CO2 + 6H2O + 6Br2 find the Percent Yield, Theoretical Yield and Actual YieldLimiting Reactant Excess Reactant

Answer 1

4C₂H₃Br₃ + 11O₂ → 8CO₂ + 6H₂O + 6Br₂ ... balanced equation

First, we must find which reactant is limiting. An easy way to do this is to divide the moles of each reactant by the corresponding coefficient in the balanced equation.

C₂H₃Br₃: 75.4 g x 1 mol / 267 g = 0.282 mols C₂H₃Br₃ (÷4 -> 0.071)

O₂: 47.1 g x 1 mol / 32 g = 1.47 mols O₂ (÷11 -> 0.134)

Therefore, C₂H₃Br₃ is limiting since 0.071 is less than 0.134.

Theoretical yield is based on the amount of limiting reactant.

Theoretical yield of CO₂ = 0.282 mols C₂H₃Br₃ x 8 mol CO2 / 4 mols C₂H₃Br₃ x 44 g CO₂/mol = 24.8 g CO₂

Find theoretical yield of the other products by the same method.

Actual yield cannot be found but must be given in the problem. You did not supply that value, so we cannot determine the percent yield which would be ...

actual yield / theoretical yield (x100%)

Excess reactant would have to be oxygen (O₂).