$C_{2}H_{2} + Br_{2} \rightarrow C_{2}H_{2}Br_{2}$
$m_{\text{tăng}}=m_{C_{2}H_{2}}=4,68 (g)$
$\Rightarrow n_{C_{2}H_{2}}=\frac{4,68}{26}=0,18 (mol)$
$n_{\text{hh}}=\frac{6,72}{22,4}=0,3 (mol)$
$\Rightarrow \%C_{2}H_{2}=\frac{0,18}{0,3}.100=60\%$
$\Rightarrow \%CH_{4}=100-60=40\%$