$n_{H_2}=8,96/22,4=0,4mol$

$PTHH :$

$Mg+H_2SO_4\to MgSO_4+H_2↑$

$Zn+H_2SO_4\to ZnSO_4+H_2↑$

$\text{a/Gọi}$ $n_{Mg}=a;n_{Zn}=b$

$\text{Ta có :}$

$m_{hh}=24a+65b=17,8g$

$n_{H_2}=a+b=0,4mol$

$\text{Ta có hpt :}$

$\left\{\begin{matrix}
24a+65b=17,8 & \\ 
a+b=0,4 & 
\end{matrix}\right.$

$⇔\left\{\begin{matrix}
a=0,2 & \\ 
b=0,2 & 
\end{matrix}\right.$

$⇒m_{Mg}=0,2.24=4,8g$

$m_{Zn}=17,8-4,8=13g$

$b/m_{muối}=m_{MgSO_4}+m_{ZnSO_4}=120×0,2+161×0,2=56,2g$

$c/n_{BaSO_4}=n_{MgSO_4}+n_{ZnSO_4}=0,2+0,2=0,4mol$

$⇒m_{BaSO_4}=0,4.232=92,8g$

\(\begin{array}{l}
a)\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
n{H_2} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
24a + 65b = 17,8\\
a + b = 0,4\\
 \Rightarrow a = b = 0,2\\
mMg = 0,2 \times 25 = 4,8g\\
mZn = 0,2 \times 65 = 13g\\
b)\\
nMgS{O_4} = nMg = 0,2\,mol\\
 \Rightarrow mMgS{O_4} = 0,2 \times 120 = 24g\\
nZnS{O_4} = nZn = 0,2\,mol\\
 \Rightarrow mZnS{O_4} = 0,2 \times 161 = 32,2g\\
c)\\
nBaS{O_4} = nZnS{O_4} + nMgS{O_4} = 0,4\,mol\\
 \Rightarrow mBaS{O_4} = 0,4 \times 232 = 92,8g
\end{array}\)