Question

When $32g$ of methane $\left(C{H}_4\right)$ and $32g$ of oxygen gas react, $20g$ of carbon dioxide is produced along with some water.What is the percent yield of carbon dioxide in this reaction?

Answer 1

Moles of Methane in $32g=\frac{32}{16}=2$ moles

Moles of Oxygen in $32g=\frac{32}{32}=1$ mole

Moles of $C{O}_2$ in $20g=\frac{20}{40}=0.5$ mole

$C{H}_4+2O_2⟶C{O}_2+2H_{2}O$

$1$ mole of Methane react with $2$ mole of $O_2$

$1$ mole of $O_2$ will react with $\frac{1}{2}$ mole of $C{H}_4=0.5$ mole of $C{H}_4$

$C{H}_4+2O_2⟶C{O}_2+2H_{2}O$

at $t=0$ $2$ $1$ - -

$2-0.5$ $1-2\times 0.5$ $0.5$ -

as upon reacting of $0.5$ mole of $C{H}_4$; $0.5$ mole of $C{O}_2$ must be produced.

The produced moles of $C{O}_2$ is $0.5$

Thus % yield is= $\frac{0.5}{0.5}\times 100=100$% .