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Trả lời:

$Mg+H_2SO_4\rightarrow MgSO_4+H_2\\\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\\2Al+3H_2SO_4\rightarrow Al_2(SO_4)_3+3H_2\\\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{3}{2}b$

$n_{H_2}=\dfrac{8,96}{22,4}=0,4\,(mol)$

Đặt $a=n_{Mg};\,b=n_{Al}$

Ta có hệ:

$\begin{cases}24a+27b=7,8\\a+\dfrac{3}{2}b=0,4\end{cases}⇔\begin{cases}a=0,1\\b=0,2\end{cases}$

$⇔\begin{cases}m_{Mg}=0,1.24=2,4\,(g)\\m_{Al}=0,2.27=5,4\,(g)\end{cases}$

Theo PT1: $n_{H_2SO_4}=n_{Mg}=0,1\,(mol)$

Theo PT2: $n_{H_2SO_4}=\dfrac{3}{2}.n_{Al}=0,3\,(mol)$

$⇒∑n_{H_2SO_4}=0,1+0,3=0,4\,(mol)$

$⇒V_{H_2SO_4}=\dfrac{n}{C_M}=\dfrac{0,4}{2}=0,2\,(l)$.