a,
Gọi hỗn hợp sau là $X$
Xét $1$ mol $CH_4$ ban đầu.
BTKL: $m_{X}=m_{CH_4}=1.16=16g$
$\overline{M}_X=6,4.2=12,8$
$\Rightarrow n_X=\dfrac{16}{12,8}=1,25(mol)$
Gọi $x$ là số mol $CH_4$ phản ứng.
$2CH_4\buildrel{{1500^oC}}\over\to C_2H_2+3H_2$
$\Rightarrow n_{C_2H_2}=0,5x (mol); n_{H_2}=1,5x (mol)$
$\Rightarrow 0,5x+1,5x+1-x=1,25$
$\Leftrightarrow x=0,25$
$\to H=\dfrac{0,25.100}{1}=25\%$
b,
$X$ chứa: $C_2H_2$ ($0,125$ mol), $H_2$ ($0,375$ mol), $CH_4$ ($0,75$ mol)
$\%m_{C_2H_2}=\dfrac{0,125.26.100}{16}=10,3125\%$
$\%m_{H_2}=\dfrac{0,375.2.100}{16}=4,6875\%$
$\Rightarrow \%m_{CH_4}=85\%$
$\%V_{C_2H_2}=\dfrac{0,125.100}{1,25}=10\%$
$\%V_{H_2}=\dfrac{0,375.100}{1,25}=30\%$
$\Rightarrow \%V_{CH_4}=60\%$