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:
$2x+3y=0,8$
$65x+27y=11,9$
$⇒x=0,1;y=0,2(mol)$
%$m_{Zn}=(0,1.65)/11,9 . 100=54,6$%
%$m_{Al}=100-54,6=45,4$%
$BTNT:n_{ZnSO_4}=0,1(mol);n_{Al_2(SO_4)_3}=0,1(mol)$
$ddY:ZnSO_4+Al_2(SO_4)_3$
$PTHH:ZnSO_4+BaCl_2→ZnCl_2+BaSO_4$
$(mol)-0,1------------0,1$
$PTHH:Al_2(SO_4)_3+3BaCl_2→2AlCl_3+3BaSO_4$
$(mol)-0,1-------------0,3$
$m_↓=233.0,4=93,2(g)$