Đặt $n_{H_2}+n_{CO}= x (mol); n_{C_4H_{10}}=y (mol)$
$n_X=\dfrac{17,92}{22,4}=0,8(mol)$
$\Rightarrow x+y=0,8$ (1)
$n_{O_2}=\dfrac{76,16}{22,4}=3,4(mol)$
$2H_2+O_2\to 2H_2O$
$C_4H_{10} + 6,5O_2\to 4CO_2+5H_2O$
$\Rightarrow 0,5x+6,5y=3,4$ (2)
(1)(2) $\Rightarrow x=0,3; y=0,5$
$\to \%V_{C_4H_{10}}=\dfrac{0,5.100}{0,8}=62,5\%$