Ban đầu CH4 là 1 mol => m = 16 gam
$2CH4 -> C2H2 + 3H2$
BĐ: $ 1mol $
PƯ : $x mol -> \frac{x}{2} mol \frac{3x}{2}mol$
Sau: $1-x mol \frac{x}{2} mol \frac{3x}{2} mol$
$\Sigma n sau = (1 + x ) mol$
BTKL
$m sau = 5\times 2 \times ( 1 + x) = 16 => x = 0,6 => H = 60$