\(\begin{array}{l}
Dat\,hoa\,tri\,M\,la\,n(n\in\mathbb{N^*})\\
Dat\,n_{Cu}=x(mol);n_M=y(mol)\\
n_{H_2}=\frac{3,136}{22,4}=0,14(mol)\\
n_{NO}=\frac{3,92}{22,4}=0,175(mol)\\
2M+2nHCl\to 2MCl_n+nH_2\\
Theo\,PT:\,y=n_M=\frac{2n_{H_2}}{n}=\frac{0,28}{n}(mol)\\
Cu^0\to Cu^{+2}+2e\\
M^0\to M^n+ne\\
N^{+5}+3e\to N^{+2}\\
Bao\,toan\,e:\,2x+yn=3n_{NO}=0,525(mol)\\
\to 2x+\frac{0,28}{n}.n=2x+0,28=0,525\\
\to x=0,1225(mol)\\
\to m_M=11,2-0,1225.64=3,36(g)\\
\to M=\dfrac{3,36}{\frac{0,28}{n}}=12n\\
\to \begin{cases} n=2\\ M_M=24(g/mol)\end{cases}\\
\to M:Mg(magie)
\end{array}\)

Đáp án:

$Mg$

Giải thích các bước giải:

Gọi hoá trị $M$ là $n\ (n∈\mathbb{N}^*).$

$n_{H_2}=\dfrac{3,136}{22,4}=0,14(mol)$

Bảo toàn electron:

$2n_{H_2}=n×n_{M}=0,28(mol)$

$n_{NO}=\dfrac{3,92}{22,4}=0,175(mol)$

Bảo toàn electron:

$n_{NO}=\dfrac{2}{3}n_{Cu}+\dfrac{1}{3}n×n_{M}=0,175(mol)\\\to n_{Cu}=\bigg(0,175-\dfrac{1}{3}×0,28\bigg)×\dfrac{3}{2}=0,1225(mol)$

$m_{M}=11,2-0,1225×64=3,36(g)$

$M_M=\dfrac{3,36}{\dfrac{0,28}{n}}=12n$

Với $n=2\to M_M=24(g/mol)$

$M$ là $Mg.$