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?

Answer 1

Assuming that the combustion formula is
CH4 + 2O2 --> 2H2O + CO2,

That means for every 1 molecule of methane(CH4) there will be one molecule of carbon dioxide(CO2) produced. Methane molecular weight 16, carbon dioxide molecular weight is 44. Then the percent yield should be:
1 * (0.374/ 16) /(0.983/44)= 0.374*44/ 0.983 * 16= 104.6%

You sure the number is correct? Percent yield should not exceed 100%