`n_{H_2}=\frac{2,24}{22,4}=0,1(mol)`
Cho $\begin{cases} Mg: x(mol)\\ Fe: y(mol)\\\end{cases}$
`24x+56y=4g(1)`
BTe:
$\mathop{Mg}\limits^{0}\to \mathop{Mg}\limits^{+2}+2e$
$\mathop{Fe}\limits^{0}\to \mathop{Fe}\limits^{+2}+2e$
$\mathop{2H}\limits^{+}+2e\to \mathop{H_2}\limits^{0}$
`2x+2y=0,2(mol)(2)`
`(1),(2)=>x=y=0,05(mol)`
Ta nhận thấy:
`n_{Mg}=n_{MgSO_4}=n_{Fe}=n_{FeSO_4}=0,05(mol)`
`CM_{MgSO_4}=CM_{FeSO_4}=\frac{0,05}{0,2}=0,25M`