How many moles of methane are required to produce $22gC{O}_2\left(g\right)$ after combustion?

Given: weight of $C{O}_2=22g$

The balanced equation for the combustion of methane is:

$C{H}_4\left(g\right)+O_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(g\right)$

By apply stoichiometry in the above equation,

$1$ mol of $C{O}_2\left(g\right)$ produced by $1$ mol of $C{H}_4\left(g\right)$

We know, $\text{mole}=\frac{\text{weight}}{\text{mol.wt.}}$

${\text{(mole)}}_{C{O}_2}=\frac{22}{44}=0.5mol$

So, $0.5$ mol of $C{O}_2\left(g\right)$ produced by $0.5$ mol of $C{H}_4\left(g\right)$

Final answer: $0.5$ mol of methane are required to produce $22gC{O}_2\left(g\right)$ after combustion.