The combustion of 0.374 kg of methane in the presence of excess oxygen produces 0.983 kg of carbon dioxide. What is the percent yield? Paired item 1)

Question

The combustion of 0.374 kg of methane in the presence of excess oxygen produces 0.983 kg of carbon dioxide. What is the percent yield?  Paired item 1)

Quizplus · Accepted Answer

The theoretical yield of carbon dioxide (CO2) from the combustion of methane (CH4) can be calculated using stoichiometry. The balanced chemical equation for the combustion of methane is: CH4 + 2O2 -> CO2 + 2H2O. From the equation, 1 mole of methane produces 1 mole of carbon dioxide. The molar mass of methane (CH4) is approximately 16 g/mol, and the molar mass of carbon dioxide (CO2) is approximately 44 g/mol. Therefore, 0.374 kg (or 374 g) of methane corresponds to 374 g / 16 g/mol = 23.375 moles of methane. Theoretically, this amount of methane would produce 23.375 moles of CO2, which is 23.375 moles * 44 g/mol = 1028.5 g or 1.0285 kg of CO2. The percent yield is calculated as (actual yield / theoretical yield) * 100 = (0.983 kg / 1.0285 kg) * 100 = 95.6%. The closest answer provided is 95.8%, which may be due to rounding differences in the calculation.

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