$n_X=\dfrac{6,72}{22,4}=0,3(mol)$
$CH\equiv CH\buildrel{{AgNO_3/NH_3}}\over\longrightarrow Ag_2C_2\downarrow$
$n_{Ag_2C_2}=\dfrac{48}{240}=0,2(mol)$
$\Rightarrow n_{C_2H_2}=0,2(mol)$
Gọi $x$, $y$ là số mol $C_3H_6$, $C_3H_8$
$\Rightarrow x+y=0,3-0,2=0,1$
$n_{H_2O}=\dfrac{6,3}{18}=0,35(mol)$
Bảo toàn $H$: $3x+4y=0,35$
Giải hệ: $x=y=0,05$
$\%V_{C_2H_2}=\dfrac{0,2.100}{0,3}=66,67\%$
$\%V_{C_3H_6}=\dfrac{0,05.100}{0,3}=16,67\%$
$\to\%V_{C_3H_8}=16,66\%$
$m_X=0,2.26+0,05.42+0,05.44=9,5g$
$\%m_{C_2H_2}=\dfrac{0,2.26.100}{9,5}=54,73\%$
$\%m_{C_3H_6}=\dfrac{0,05.42.100}{9,5}=22,11\%$
$\to\%m_{C_3H_8}=23,16\%$