$2Mg+O_2\xrightarrow{{t^o}} 2MgO$

$4Al+3O_2\xrightarrow{{t^o}} 2Al_2O_3$

$2Zn+O_2\xrightarrow{{t^o}} 2ZnO$

Bảo toàn khối lượng: $m_{O_2}=m+9,6-m=9,6g$

$\to n_{O_2}=\dfrac{9,6}{32}=0,3(mol)$

Ta có: $n_{O_2}=\dfrac{1}{2}n_{Mg}+\dfrac{3}{4}n_{Al}+\dfrac{1}{2}n_{Zn}$

$\to 4n_{O_2}=2n_{Mg}+3n_{Al}+2n_{Zn}=0,3.4=1,2(mol)$

$Mg+H_2SO_4\to MgSO_4+H_2$

$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$

$Zn+H_2SO_4\to ZnSO_4+H_2$

Ta có: $n_{H_2}=n_{Mg}+\dfrac{3}{2}n_{Al}+n_{Zn}$

$\to 2n_{H_2}=2n_{Mg}+3n_{Al}+2n_{Zn}$

TN2 dùng gấp đôi kim loại TN1 nên:

$2n_{H_2}=2.1,2$

$\to n_{H_2}=1,2(mol)$

$\to V=1,2.22,4=26,88l$

Theo PTHH, $n_{H_2SO_4}=n_{H_2}=1,2(mol)$

$\to m_{H_2SO_4}=1,2.98=117,6g$