$n_{\ hh \ khí}=\dfrac{6,72}{22,4}=0,3(mol)$
Cho $A\ \begin{cases}CH_4:x(mol)\\C_2H_2:y(mol)\\\end{cases}$
$n_{\downarrow}=\dfrac{40}{100}=0,4(mol)$
Ta có: $x+y=0,3(mol)(1)$
Bảo toàn $C$:
$\to n_{CO_2}=x+2y(mol)$
$Ca(OH)_2+CO_2\to CaCO_3\downarrow+H_2O$
$n_{CaCO_3}=n_{CO_2}=0,4(mol)$
$\to x+2y=0,4(mol)(2)$
Từ $(1),(2)\to \begin{cases}x=0,2(mol)\\y=0,1(mol)\\\end{cases}$
$\to V_{CH_4\ trong \ A}=0,2.22,4=4,48(l)$