a) PTHH
$C_2H_2+2Br_2→C_2H_2Br_4$
$CH_4$ không tác dụng được với nước $B_2$
b) $n_{Br_2}=\dfrac{16}{160}=0,1(mol)$
$n_{\text{hh khí}}=\dfrac{3,36}{22,4}=0,15(mol)$
$n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=\dfrac{1}{2}×0,1=0,05(mol)$
%$V_{C_2H_2}=\dfrac{0,05}{0,15}×100$% $=$ $33,33$%
→ %$V_{CH_4}=100$% $-$ $33,33$% $=$ $66,67$%