$\rm dd\ X :$ $\begin{cases} \rm MgCl_2 : a(mol)\\\rm FeCl_3 : b(mol)\\\rm CuCl_2 : c(mol) \end{cases}$

Thí nghiệm 1:

$\rm 2FeCl_3+H_2S\to 2FeCl_2+S\downarrow+2HCl$

$\rm CuCl_2+H_2S\to CuS\downarrow+2HCl$

$\rm 16(g)\ kết\ tủa :$ $\begin{cases} \rm S : 0,5b(mol)\\\rm CuS : c(mol) \end{cases}$

$\rm m_{kt}=32.0,5b+96b=16(g)\ (1)$

Thí nghiệm 2:

$\rm Na_2S+MgCl_2+2H_2O\to Mg(OH)_2\downarrow+2NaCl+H_2O$

$\rm 3Na_2S+2FeCl_3\to 2FeS\downarrow+S\downarrow+6NaCl$

$\rm Na_2S+CuCl_2\to CuS\downarrow+2NaCl$

$\rm 30,6(g)\ kết\ tủa :$ $\begin{cases} \rm Mg(OH)_2 : a(mol)\\\rm FeS : b(mol)\\ \rm S : 0,5b(mol)\\ \rm CuS : c(mol) \end{cases}$

$\rm m_{kt}=58a+88b+32.0,5b+96c=30,6(g)\ (2)$

Thí nghiệm 3:

$\rm Na_2S+MgCl_2+2H_2O\to Mg(OH)_2\downarrow+2NaCl+H_2O$

$\rm Na_2S+FeCl_2\to FeS\downarrow+2NaCl$

$\rm Na_2S+CuCl_2\to CuS\downarrow+2NaCl$

$\rm 29(g)\ kết\ tủa :$ $\begin{cases} \rm Mg(OH)_2 : a(mol)\\\rm FeS : b(mol)\\ \rm CuS : c(mol) \end{cases}$

$\rm m_{kt}=58a+88b+96c=29(g)\ (3)$

`(1)(2)(3)=>` $\rm a=b=0,1(mol)\ , c=0,15(mol)$

$\rm m_{MgCl_2}=0,1.95=9,5(g)$

$\rm m_{FeCl_3}=0,1.162,5=16,25(g)$

$\rm m_{CuCl_2}=0,15.135=20,25(g)$

$\bullet$ Xét thí nghiệm 1: Gọi $\begin{cases} \rm n_{MgCl_2}=a\ (mol)\\\rm n_{FeCl_3}\:=b\ (mol)\\\rm n_{CuCl_2}\;=c\ (mol)\end{cases}$

Phản ứng xảy ra:

$\rm\quad\quad\;\;\; H_2S+2FeCl_3\to 2FeCl_2+S+2HCl\\(mol)\hspace{2,3cm}b\hspace{1,1cm}\to\hspace{1,1cm}0,5b$

$\rm\quad\quad\;\;\; H_2S+CuCl_2\to CuS+2HCl\\(mol)\hspace{2,2cm}b\hspace{,4cm}\to\hspace{,4cm}b$

$\rm ⇒m_↓=16b+96c=16\ (1)$

$\bullet$ Xét thí nghiệm 2: 

Phản ứng xảy ra:

$\rm\quad\quad\;\;\; MgCl_2+Na_2S+2H_2O\to Mg(OH)_2+2NaCl+H_2O\\(mol)\hspace{0,8cm}a\hspace{2,4cm}\to\hspace{2,4cm}a$

$\rm\quad\quad\;\;\; 3Na_2S+2FeCl_3\to 2FeS+S+6NaCl\\(mol)\hspace{2,7cm}b\hspace{0,5cm}\to\hspace{0,5cm}b\hspace{0,6cm}0,5b$

$\rm\quad\quad\;\;\;Na_2S + CuCl_2 → CuS + 2NaCl\\(mol)\hspace{2,3cm}c\hspace{0,4cm}\to\hspace{0,5cm}c$

$\rm ⇒m_↓=58a+104b+96c=30,6\ (2)$

$\bullet$ Xét thí nghiệm 3: Gọi $\begin{cases} \rm n_{MgCl_2}=a\ (mol)\\\rm n_{FeCl_2}\:=b\ (mol)\\\rm n_{CuCl_2}\;=c\ (mol)\end{cases}$

Phản ứng xảy ra:

$\rm\quad\quad\;\;\; MgCl_2+Na_2S+2H_2O\to Mg(OH)_2+2NaCl+H_2O\\(mol)\hspace{0,8cm}a\hspace{2,4cm}\to\hspace{2,4cm}a$

$\rm\quad\quad\;\;\;Na_2S + FeCl_2 → FeS + 2NaCl\\(mol)\hspace{2,3cm}b\hspace{0,4cm}\to\hspace{0,5cm}b$

$\rm\quad\quad\;\;\;Na_2S + CuCl_2 → CuS + 2NaCl\\(mol)\hspace{2,3cm}c\hspace{0,4cm}\to\hspace{0,5cm}c$

$\rm ⇒m_↓=58a+88b+96c=29\ (3)$

Từ $(1),(2)$ và $(3)$ ta giải được: $\rm a=b=0,1; c=0,15$

Vậy $\rm m_{MgCl_2}=0,1×95=9,5\ (gam)$

$\rm\quad\;\; m_{FeCl_3}=0,1×162,5=16,25\ (gam)$

$\rm\quad\;\; m_{CuCl_2}=0,15×135=20,25\ (gam)$