Câu `4:`
`n_{SO_2}=\frac{7,616}{22,4}=0,34(mol)`
`n_S=\frac{0,64}{32}=0,02(mol)`
Cho $11,9(g) \begin{cases} Al: x(mol)\\ Zn: y(mol)\\\end{cases}$
`27x+65y=11,9(g)(1)`
BTe:
$\mathop{Al}\limits^{0}\to \mathop{Al}\limits^{+3}+3e$
$\mathop{Zn}\limits^{0}\to \mathop{Zn}\limits^{+2}+2e$
$\mathop{S}\limits^{+6}+2e\to \mathop{S}\limits^{+4}$
$\mathop{S}\limits^{+6}+6e\to \mathop{S}\limits^{0}$
Ta có: `3x+2y=2n_{SO_2}+6n_S`
`=> 3x+2y=0,8(mol)(2)`
`(1),(2)=>x=0,2(mol), y=0,1(mol)`
Bảo toàn nguyên tố:
`n_{Al_2(SO_4)_3}=0,5n_{Al}=0,1(mol)`
`n_{ZnSO_4}=n_{Zn}=0,1(mol)`
`m_{\text{muối}}=m_{Al_2(SO_4)_3}+m_{ZnSO_4}`
`=> m_{\text{muối}}=0,1.(342+161)=50,3g`
