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%

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We first assume that this gas is an ideal gas where it follows the ideal gas equation. the said equation is expressed as: pv = nrt. from this equation, we can predict the changes in the pressure, volume and temperature. if the volume and the temperature of this gas is doubled, then the pressure still stays the same.