4.00 mol C 8 H 18 reacts with 40.0 mol O 2 , in actual forming 917 g CO2. What is the percentage yield of such reaction?

% Yield = 81.4%

Explanation:

Based on the reaction:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O

1 mole of C₈H₁₈ reacts with 25/2 moles of O₂ to produce 8 moles of CO₂

To solve this question we must, as first, find limiting reactant. With limiting reactant we can find theoretical yield of CO₂. Percent yield is defined as:

% Yield = Actual yield (917g CO₂) / Theoretical Yield * 100

For a complete reaction of 4.00 moles of C₈H₁₈ are required:

4.00moles C₈H₁₈ * (25/2 moles O₂ / 1mol C₈H₁₈) = 50.0 moles of O₂

As there are just 40.0 moles, O₂ is limiting reactant

The theoretical yield of CO₂ from 40.0 moles of O₂ is:

40.0 moles O₂ * (8 moles CO₂ / 25/2 moles O₂) = 25.6 moles of CO₂

As molar mass of CO₂ is 44.01g/mol:

25.6 moles of CO₂ * (44.01g / mol) = 1126.7g of CO₂ is theoretical yield

And percent yield is:

% Yield = 917g / 1126.7g * 100

pv=nrt

explanation: