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%

Adozen is 1221/12 = 1.751.75 x 12 = 21therefore, 1.75 is the answer.