in a complete combustion reaction, 72.15g of pentane, c5h12(g) reacts with 300 g of oxygen

Question

write the balanced chemical equation for this reaction

 

identify the limiting reactant

calculate the mass of carbon dioxide produced

determine the actual yield of carbon dioxide if the percentage yield was 62.9%

Answer 1

Complete combustion of pentane

C₅H₁₂ + 8O₂ ==> 5CO₂ + 6H₂O ... balanced equation

moles C5H12 = 72.15 g x 1 mole/72 g = 1.00 moles

moles O2 = 300 g x 1 mole/32 g = 9.375 moles

Find limiting reactant by dividing moles by coefficient:  pentane: 1 mol/1 = 1;  O₂ 9.375 mole/8 = 1.17.  So, pentane is limiting.

 

Mass of CO₂ produced:  1.00 moles C₅H₁₂ x 5 mole CO₂/1 mole C₅H₁₂ = 5 moles CO₂ x 44 g/mole = 220 g CO₂ produced

 

Actual yield = theoretical yield x % yield = 220 g x 0.629 = 138 g