Sơ đồ phản ứng:
$\rm 9,9 (g) \begin{cases} \rm M(II)\\\rm R(III)\\\end{cases} \xrightarrow{\rm + HNO_3} \rm \ dd \ A +\ 0,5 (mol) \begin{cases}\rm NO_2: x (mol)\\ \rm NO: y(mol)\\\end{cases}$
`∑M_B=\frac{m_{NO_2}+m_{NO}}{n_{NO_2}+n_{NO}}`
`=> 19,8.2=\frac{46x+30y}{0,5}`
`=> 46x+30y=19,8g(1)`
Mặc khác:
`x+y=0,5(2)`
`(1),(2)=> ` $\begin{cases}x=0,3(mol)\\ y=0,2(mol)\\\end{cases}$
BTe:
$\mathop{M}\limits^{0}\to \mathop{M}\limits^{+2}+2e$
$\mathop{R}\limits^{0}\to \mathop{R}\limits^{+3}+3e$
$\mathop{N}\limits^{+5}+3e\to \mathop{N}\limits^{+2}$
$\mathop{N}\limits^{+5}+e\to \mathop{N}\limits^{+4}$
`=> 2n_{M}+3n_{R}=3n_{NO}+n_{NO_2}=0,9(mol)=n_{HNO_3}`
`=> n_{NO_3^{-}}=0,9(mol)`
`m_{\text{muối}}=m_{\text{kl}}+m_{NO_3^{-}}`
`=> m_{\text{muối}}=9,9+0,9.62=65,7g`
