Câu 1:

Đặt hóa trị của R là $n(n>0)$

$a)PTHH:2R+2nHCl\to 2RCl_n+nH_2$

Theo PT: $n_R=n_{RCl_n}$

Ta có $m_{dd_A}=m_R+m_{dd_{HCl}}-m_{H_2}$

$\Rightarrow 1,2+200-m_{H_2}=201,1$

$\Rightarrow m_{H_2}=0,1(g)$

$\Rightarrow n_{H_2}=\dfrac{0,1}{2}=0,05(mol)$

Theo PT: $n_R.n=2.n_{H_2}=0,1(mol)$

$\Rightarrow \dfrac{1,2.n}{M_R}=0,1$

$\Rightarrow M_R=12n$

Với $n=1\Rightarrow M_R=12(loại)$

Với $n=2\Rightarrow M_R=24(nhận)$

Vậy R là magie $(Mg)$

$b)PTHH:Mg+2HCl\to MgCl_2+H_2$

Theo PT: $n_{HCl}=2n_{H_2}=0,1(mol);n_{MgCl_2}=n_{H_2}=0,05(mol)$

$\Rightarrow C\%_{HCl}=\dfrac{0,1.36,5}{200}.100\%=1,825\%$

$m_{MgCl_2}=0,05.95=4,75(g)$

$\Rightarrow C\%_{MgCl_2}=\dfrac{4,75}{1,2+200-0,1}.100\%\approx 2,36\%$

Câu 2:

Đặt KL trung bình là $\overline{R}\to R(II)$

$n_{H_2}=\dfrac{0,672}{22,4}=0,03(mol)$

$\Rightarrow PTHH:\overline{R}+2HCl\to \overline{R}Cl_2+H_2$

Theo PT: $n_{\overline{R}}=n_{H_2}=0,03(mol)$

$\Rightarrow M_{\overline{R}}=\dfrac{1,7}{0,03}=56,67(g/mol)$

$\Rightarrow M_{\overline{R}}<56,67<M_{Zn}=65$

$\Rightarrow M_A<65(1)$

$PTHH:A+2HCl\to ACl_2+H_2$

Đổi $200ml=0,2l\Rightarrow n_{HCl}=0,5.0,2=0,1(mol)$

Theo PT: $n_{HCl}<2.n_A=0,1$

$\Rightarrow n_A<0,05$

$\Rightarrow M_A>\dfrac{1,9}{0,05}=38(2)$

Từ $(1)(2)\Rightarrow M_A=40(g/mol)$

Vậy A là canxi $(Ca)$