The combustion of butane produced 0.0243 kg of carbon dioxide. What mass of butane is required if the yield is 89 %. Paired item 1)

Question

The combustion of butane produced 0.0243 kg of carbon dioxide. What mass of butane is required if the yield is 89 %.  Paired item 1)

Quizplus · Accepted Answer

The balanced equation for the combustion of butane (C4H10) is: 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O. From the equation, 2 moles of butane produce 8 moles of CO2. The molar mass of CO2 is 44.01 g/mol, so 0.0243 kg (24.3 g) of CO2 is 24.3 g / 44.01 g/mol = 0.552 mol of CO2. Since 8 moles of CO2 come from 2 moles of butane, 0.552 mol of CO2 comes from 0.138 mol of butane. The molar mass of butane is 58.12 g/mol, so 0.138 mol of butane is 0.138 mol * 58.12 g/mol = 8.02 g. Considering the yield is 89%, the actual mass of butane required is 8.02 g / 0.89 = 9.01 g.

The Combustion of Butane Produced 0