Đáp án:

$a)\ R:\ Mg\\b)\ \%m_{Mg}=30\%;\ \%m_{Cu}=70\%$

Giải thích các bước giải:

$a)$

Thí nghiệm 1:

$2R+2nHCl\to 2RCl_n+nH_2$

$n_{H_2}=\dfrac{3,136}{22,4}=0,14(mol)\\\to n_{R}=\dfrac{2n_{H_2}}{n}=\dfrac{0,28}{n}(mol)\\\to n×n_R=0,28(mol)$

Thí nghiệm 2:

Khối lượng ở thí nghiệm 2 gấp đôi thí nghiệm 1, nên số mol cũng gấp đôi.

$4R+nO_2\xrightarrow{t^o}2R_2O_n$

$2Cu+O_2\xrightarrow{t^o}2CuO$

$n_{O_2}=\dfrac{1}{2}n_{Cu}+\dfrac{2}{4}n×n_R=\dfrac{5,88}{22,4}=0,2625(mol)$

$\to n_{Cu}=0,245(mol)$

$\to m_{R}=22,4-0,245×64=6,72(g)$

$M_R=\dfrac{6,72}{\dfrac{0,28×2}{n}}=12n(g/mol)$

Với $n=2\to M_R=24(g/mol)$

Vậy $R$ là $Mg.$

$b)$

$Mg+2HCl\to MgCl_2+H_2$

$n_{H_2}=n_{Mg}=0,14(mol)$

$\%m_{Mg}=\dfrac{0,14×24}{11,2}×100\%=30\%$

$\%m_{Cu}=100\%-30\%=70\%$

\(\begin{array}{l}
a)\\
Dat\,hoa\,tri\,R\,la\,n(n\in\mathbb{N^*})\\
Trong\,11,2g\,X\,dat\,n_{Cu}=x(mol);n_R=y(mol)\\
\to Trong\,22,4g\,X\,co\,n_{Cu}=2x(mol);n_R=2y(mol)\\
n_{H_2}=\frac{3,136}{22,4}=0,14(mol)\\
n_{O_2}=\frac{5,88}{22,4}=0,2625(mol)\\
2R+2nHCl\to 2RCl_n+nH_2\\
Theo\,PT:\,y=n_R=\frac{2n_{H_2}}{n}=\frac{0,28}{n}(mol)\\
4R+nO_2\xrightarrow{t^o}2R_2O_n\\
2Cu+O_2\xrightarrow{t^o}2CuO\\
Theo\,PT:\,\frac{n.n_R}{4}+\frac{1}{2}n_{Cu}=n_{O_2}\\
\to \frac{n.0,28.2}{4.n}+0,5.2x=0,2625\\
\to 0,14+x=0,2625\\
\to x=0,1225(mol)\\
\to m_R=11,2-0,1225.64=3,36(g)\\
\to M_R=\dfrac{3,36}{\frac{0,28}{n}}=12n\\
\to \begin{cases} n=2\\ M_R=24(g/mol) \end{cases}\\
\to R:Magie(Mg)\\
b)\\
x=n_{Cu}=0,1225(mol)\\
\%m_{Cu}=\frac{0,1225.64}{11,2}.100\%=70\%\\
\%m_{Mg}=100-70=30\%
\end{array}\)