The combustion of propane produced 8.53 g of carbon dioxide. What mass of propane is required if the yield is 93 %.

Question

Quizplus · Accepted Answer

The combustion of propane (C3H8) can be represented by the balanced chemical equation: C3H8 + 5O2 -> 3CO2 + 4H2O. From this equation, 1 mole of propane produces 3 moles of CO2. The molar mass of CO2 is 44.01 g/mol, so 8.53 g of CO2 corresponds to 8.53 g / 44.01 g/mol = 0.1938 moles of CO2. Since 3 moles of CO2 come from 1 mole of propane, 0.1938 moles of CO2 come from 0.1938 / 3 = 0.0646 moles of propane. The molar mass of propane is approximately 44.1 g/mol, so the mass of propane is 0.0646 moles * 44.1 g/mol = 2.85 g. Considering the yield is 93%, the actual mass of propane required is 2.85 g / 0.93 = 3.06 g.

The Combustion of Propane Produced 8