The combustion of 0.136 kg of methane in the presence of excess oxygen produced 353 g of carbon dioxide what is the percentage yield

The combustion reaction for methane is as follows

CH4 + 2O2 - > CO2 + 2H2O

Stoichiometry of methane to CO2 is 1:1

Since oxygen is in excess, methane is the limiting reactant therefore amount of CO2 produced depends on methane reacted,

Number of methane moles reacted = 136 g / 16 g/mol = 8.5 mol

Therefore moles of CO2 to be produced = 8.5 mol

Theoretically mass of CO2 to be produced = 8.5 mol x 44 g/mol = 374 g

However actual yield = 353 g

Percent yield = actual yield / theoretical yield x 100%

Percent yield = 353 g / 374 g x100%

Percentage yield of CO2 = 94.4%