Bạn tham khảo:
$a/$
$Zn+H_2SO_4 \to ZnSO_4+H_2$
$Fe+H_2SO_4 \to FeSO_4+H_2$
$b/$
$n_{H_2}=0,2(mol)$
$n_{Zn}=a(mol)$
$n_{Fe}=b(mol)$
$m_{hh}=65a+56b=12,1(1)$
$n_{H_2}=a+b=0,2(2)$
$(1)(2)$
$a=b=0,1$
$m_{Zn}=0,1.65=6,5(g)$
$m_{Fe}=5,6(g)$
$c/$
$n_{H_2SO_4}=0,1+0,1=0,2(mol)$
$m_{H_2SO_4}=0,2.98=19,6(g)$