Cho $\begin{cases} Fe: x(mol)\\ Al: y(mol)\\\end{cases}$
`56x+27y=13,9g(1)`
`n_{H_2}=\frac{7,84}{22,4}=0,35(mol)`
BTe:
$\mathop{Fe}\limits^{0}\to \mathop{Fe}\limits^{+2}+2e$
$\mathop{Al}\limits^{0}\to \mathop{Al}\limits^{+3}+3e$
$\mathop{2H}^{+}2e\to \mathop{H_2}\limits^{0}$
`=> 2x+3y=0,7(mol)(2)`
`(1),(2)=>x=0,2(mol), y=0,1(mol)`
`a) ` `%m_{Fe}=\frac{56.0,2.100%}{13,9}\approx80,57%`
`%m_{Al}=\frac{27.0,1.100%}{13,9}\approx 19,42%`
`b)` `n_{H_2SO_4}=n_{H_2}=0,35(mol)`
`CM_{H_2SO_4}=\frac{n}{V}=\frac{0,35}{0,1}=3,5M`
`c)` BT nguyên tố:
`n_{Fe}=n_{FeSO_4}=0,2(mol)`
`=> m_{FeSO_4}=0,2.152=30,4g`
`n_{Al_2(SO_4)_3}=0,5n_{Al}=0,05(mol)`
`=> m_{Al_2(SO_4)_3}=342.0,05=17,1g`