- Xét TN2:
$n_{SO_4^{2-}\text{muối}}=\dfrac{3,5a-a}{96}=\dfrac{5a}{192}(mol)$
$n_{SO_2}=\dfrac{10,08}{22,4}=0,45(mol)$
Ta có: $n_{SO_4^{2-}\text{muối}}=\dfrac{n_e}{2}=n_{SO_2}$
$\to \dfrac{5a}{192}=0,45$
$\to a=17,28$
- Xét TN1:
Đặt $n_{H_2}=x(mol)$
Bảo toàn hidro: $n_{H_2SO_4}=n_{H_2}=x(mol)$
BTKL:
$17,28+98x=3.17,28+2x$
$\to x=0,36$
Vậy $V=8,064l$