$n_{H_2}=5,6/22,4=0,25mol$
$PTHH :$
$Zn+H_2SO_4\to ZnSO_4+H_2↑$
$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2↑$
Gọi $n_{Zn}=a;n_{Al}=b$
$\text{Ta có :}$
$m_{hh}=65a+27b=9,2g$
$n_{H_2}=a+1,5b=0,25mol$
$\text{Ta có hpt :}$
$\left\{\begin{matrix}
65a+27b=9,2 & \\
a+1,5b=0,25 &
\end{matrix}\right.$
$⇔\left\{\begin{matrix}
a=0,1 & \\
b=0,1 &
\end{matrix}\right.$
$⇒\%m_{Zn}=\dfrac{0,1.65.100\%}{9,2}=70,65\%$
$\%m_{Al}=100\%-70,65\%=29,35\%$
$b/PTHH :$
$Zn+H_2SO_4\to 2H_2O+SO_2↑+ZnSO_4$
$2Al+6H_2SO_4\to Al_2(SO_4)_3+6H_2O+2SO_2↑$
$V_{SO_2}=(0,1+0,1).22,4=4,48l$
