Sơ đồ phản ứng:
$\rm 18,2(g)\ A \begin{cases}\rm \ Al: a \\ \rm Cu: b\\\end{cases} \xrightarrow[\rm 0,2 \ (mol)\ HNO_3]{\rm \ 1,2 (mol)\ H_2SO_4} \begin{cases} \rm Al^{3+}\\ \rm \ Cu^{2+}\\ \rm SO_4^{2-}\\ \rm NO_3^{-}\\ \rm H^{+} \ dư \\\end{cases}\rm \ + H_2O\ +\ 0,4 (mol) \begin{cases}\rm NO: x\\ \rm SO_2: y\\\end{cases}$
Ta có:
`x+y=0,4(mol)(1)`
Mặc khác:
`∑_{D}=\frac{m_{NO}+m_{SO_2}}{n_{NO}+n_{SO_2}}`
`=> 23,5.2=\frac{30x+64y}{0,4}`
`=> 30x+64y=18,8g(2)`
`(1),(2)=> ` $\begin{cases}x=0,2(mol)\\ y=0,2(mol)\\\end{cases}$
BTe:
$\mathop{Al}\limits^{0}\to \mathop{Al}\limits^{+3}+3e$
$\mathop{Cu}\limits^{0}\to \mathop{Cu}\limits^{+2}+2e$
$\mathop{N}\limits^{+5}+3e\to \mathop{N}\limits^{+2}$
$\mathop{S}\limits^{+6}+2e\to \mathop{S}\limits^{+4}$
`=> 3a+2b=1(mol)(3)`
Mặc khác: `27a+64b=18,2g(4)`
`(3),(4)=>` $\begin{cases}x=0,2(mol)\\ y=0,2(mol)\\\end{cases}$
`=> m_{\text{Al trong A}}=0,2.27=5,4g`
`n_{Al^{3+}}=n_{Al}=n_{Cu^{2+}}=0,2(mol)`
`n_{SO_4^{2-}}=n_{H_2SO_4}-n_{SO_2}=1,2-0,2=1(mol)`
`n_{NO_3^{-}}=n_{HNO_3}-n_{NO}=0(mol)`
`=>` Trong `C` không có `NO_3^{-}`
Bảo toàn điện tích:
`n_{H^{+}\text{ dư}}=2n_{SO_4^{2-}}-(3n_{Al^{3+}}+2n_{Cu^{2+}})=1(mol)`
`m_{\text{ chất tan trong C}}=m_{Al^{3+}}+m_{Cu^{2+}}+m_{SO_4^{2-}}+m_{H^{+} \text{dư}}`
`=> m=27.0,2+64.0,2+1.96+1=115,2g`
