Gọi $a$, $b$, $c$ (ml) là thể tích $H_2$, $N_2$, $CH_4$

$\Rightarrow a+b+c=110$              $(1)$

$104ml$ khí gồm $CO_2$, $N_2$, $O_2$ dư nếu có 

$32ml$ khí gồm $N_2$, $O_2$ dư nếu có

$\Rightarrow V_{CO_2}=104-32=72ml$

$2H_2+O_2\buildrel{{t^o}}\over\to 2H_2O$

$CH_4+2O_2\buildrel{{t^o}}\over\to CO_2+2H_2O$

$\Rightarrow c=72$                        $(2)$

$V_{O_2\text{dư}}=32-b (ml)$

$\Rightarrow V_{O_2\text{pứ}}=180-(32-b)=b+148(ml)$

Theo PTHH, $\dfrac{1}{2}V_{H_2}+2V_{CH_4}=V_{O_2}$

$\Rightarrow 0,5a+2c=b+148$

$\Leftrightarrow 0,5a-b+2c=148$  $(3)$

Từ $(1)(2)(3)$ suy ra $a=28; b=10; c=72$

$V_{H_2}=28ml$

$V_{N_2}=10ml$

$V_{CH_4}=72ml$