`n_{H_2}=\frac{5,6}{22,4}=0,25(mol)`
Cho `Al` và `Fe` lần lượt là `x,y` mol.
Ta có: `27x+56y=8,3g(1)`
Bảo toàn `e`.
$\mathop{Al}\limits^{+0} \to \mathop{Al}\limits^{+3}+3e$ `||` $\mathop{2H}\limits^{+1}+2e \to \mathop{H_2}\limits^{+0}$
$\\ \mathop{Fe}\limits^{+0} \to \mathop{Fe}\limits^{+2}+2e$
`=> n_{H_2}=1,5x+y=0,25(mol)(2)`
`(1),(2)=> x=y=0,1(mol)`
`=> %m_{Al}=\frac{27.0,1.100%}{8,3}\approx 32,53%`
`=> %m_{Fe}=\frac{56.0,1.100%}{8,3}\approx 67,45%`