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What mass of O2 is required to produce 14.5 g of CO2 if the reaction has a 65.0 % yield? CH4 + O2 ® CO2 + H2O
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First balance the reaction: \[CH_{4}+2O_{2} \rightarrow CO_{2}+2H_{2}O\] Now let's find moles: \[n_{CO_{2}}=\frac{14.5}{44}=0.330\] Since the reaction has a 65% yield, 0.330 moles is 65% of what we would expect to get had everything reacted to completion. In other words, 0.330/x = 0.65, where x is the theoretical number of moles of CO2 we would expect. In this case, x = 0.508. Now use 0.508 moles of CO2, and use stoichiometry to find moles of O2 required, then convert that to mass and you're done!
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