Đặt $n_{Zn}=x(mol);n_{Al}=y(mol)$

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

$Zn^o→Zn^{+2}+2e$

$x------2x(mol)$

$Al^o→Al^{+3}+3e$

$y------3y(mol)$

$H_2^--→H_2^o+2e$

$0,4--0,4--0,8(mol)$

$BTe:⇒2x+3y=0,8$

Theo đề bài ta có hpt:

$2x+3y=0,8$

$65x+27y=11,9$

$⇒x=0,1;y=0,2(mol)$ 

%$m_{Zn}=(0,1.65)/11,9 . 100=54,6$%

%$m_{Al}=100-54,6=45,4$%

$BTNT:n_{ZnSO_4}=0,1(mol);n_{Al_2(SO_4)_3}=0,1(mol)$

$ddY:ZnSO_4+Al_2(SO_4)_3$

$PTHH:ZnSO_4+BaCl_2→ZnCl_2+BaSO_4$

$(mol)-0,1------------0,1$

$PTHH:Al_2(SO_4)_3+3BaCl_2→2AlCl_3+3BaSO_4$

$(mol)-0,1-------------0,3$

$m_↓=233.0,4=93,2(g)$

$\text{a)}\\Zn + H_2SO_4 \to ZnSO_4 + H_2\\\hspace{0,2cm}a\hspace{5,2cm}a\\2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\\hspace{0,2cm}b\hspace{6,1cm}1,5b\\m_{Zn,\,Al}=11,9\,(g)\\\to 65a+27b=11,9\,(1)\\n_{H_2}=\dfrac{8,96}{22,4}=0,4\,(mol)\\\to a+1,5b=0,4\,(2)\\\text{Từ (1), (2)}\to \begin{cases}65a+27b=11,9\\a+1,5b=0,4\end{cases}\to \begin{cases}a=0,1\\b=0,2\end{cases}\\\to \%Zn =\dfrac{0,1.65}{11,9}.100=54,62\%\\\to \%Al=100-54,62=45,38\%\\\text{b)}\\ZnSO_4 + BaCl_2 \to BaSO_4 + ZnCl_2\,\,\,(1)\\\hspace{0,4cm}0,1\hspace{3,4cm}0,1\\Al_2(SO_4)_3 + 3BaCl_2 \to 3BaSO_4 + 2AlCl_3\,\,\,(2)\\\hspace{0,4cm}0,1\hspace{4,5cm}0,3\\\to m=m_{BaSO_4\,(1)}+m_{BaSO_4\,(2)}\\\hspace{1cm}=0,1.233+0,3.233\\\hspace{1cm}=93,2\,(g)$