Gọi $n_{Al} = a(mol) ; n_{Zn} = b(mol) \Rightarrow 27a + 65b = 15,15(1)$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$Zn + 2HCl \to ZnCl_2 + H_2$
Theo PTHH :
$n_{H_2} = 1,5a + b = \dfrac{10,08}{22,4} = 0,45(2)$
Từ (1)(2) suy ra a = 0,2 ; b = 0,15
$\%m_{Zn} = \dfrac{0,15.65}{15,15}.100\% = 64,36\%$