Fe + 2HCl → FeCl2 + H2 (1)
2Al + 6HCl → 2AlCl3 + 3H2 (2)
\(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)
Gọi số mol của Fe và Al lần lượt là \(x,y\)
Ta có: \(\left\{{}\begin{matrix}x+\dfrac{2}{3}y=0,45\\56x+27y=13,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,15\left(mol\right)\\y=0,2\left(mol\right)\end{matrix}\right.\)
2Fe + 6H2SO4 → Fe2(SO4)3 + 3SO2 + 6H2O (3)
2Al + 6H2SO4 → Al2(SO4)3 + 3SO2 + 6H2O (4)
Theo PT3: \(n_{SO_2}=\dfrac{3}{2}n_{Fe}=\dfrac{3}{2}\times0,15=0,225\left(mol\right)\)
Theo PT4: \(n_{SO_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}\times0,2=0,3\left(mol\right)\)
\(\Rightarrow\Sigma n_{SO_2}=0,225+0,3=0,525\left(mol\right)\)
\(\Rightarrow V_{SO_2}=0,525\times22,4=11,76\left(l\right)\)
