Đáp án + Giải thích các bước giải:

$a\bigg)$

$n_{H_2}=\frac{5,6}{22,4}=0,25(mol)$

Đặt $n_{Fe}=x(mol);n_{Mg}=y(mol)$

$\to 56x+24y=9,2(1)$

$Fe+2HCl\to FeCl_2+H_2$

$Mg+2HCl\to MgCl_2+H_2$

Theo PT: $\sum n_{H_2}=n_{Fe}+n_{Mg}$

$\to x+y=0,25(2)$

Từ $(1)(2)\to x=0,1(mol);y=0,15(mol)$

$\to \begin{cases} m_{Fe}=0,1.56=5,6(g)\\ m_{Mg}=0,15.24=3,6(g) \end{cases}$

$b\bigg)$

Đổi $600ml=0,6l$

$\to n_{HCl}=0,6.1=0,6(mol)$

$\to n_{HCl(dư)}=0,6-0,25.2=0,1(mol)$

$B:HCl(dư),MgCl_2,FeCl_2$

Theo PT: $n_{MgCl_2}=0,15(mol);n_{FeCl_2}=0,1(mol)$

$\to m_{muối\,khan}=m_{MgCl_2}+m_{FeCl_2}$

$\to m_{muối\,khan}=0,15.95+0,1.127=26,95(g)$

$c\bigg)$

$n_{Cl^-}=n_{HCl(dư)}+2n_{MgCl_2}+2n_{FeCl_3}=0,6(mol)$

$Ag^++Cl^-\to AgCl\downarrow$

$\to n_{AgCl}=n_{Cl^-}=0,6(mol)$

$\to m_{\downarrow}=m_{AgCl}=0,6.143,5=86,1(g)$