1, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Gọi: \(\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\) ⇒ 56x + 27y = 22 (1)
Ta có: \(n_{H_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Fe}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y=0,8\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,4\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,2.56}{22}.100\%\approx50,91\%\\\%m_{Al}\approx49,09\%\end{matrix}\right.\)
2, Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,8\left(mol\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,8.98}{250}.100\%=31,36\%\)