Đáp án:
Giải thích các bước giải:
$n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)$
$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
$n_{Al} = \dfrac{2}{3}n_{H_2} = 0,1(mol)$
$n_{Al_2(SO_4)_3} = n_{Al_2(SO_4)_3.18H_2O} = \dfrac{99,9}{666} =0,15(mol)$
$Al_2O_3 + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2O$
Theo PTHH :
$n_{Al_2(SO_4)_3} = 0,5n_{Al} + n_{Al_2O_3}$
$\to n_{Al_2O_3} = 0,15 - 0,1.0,5 = 0,1(mol)$
Suy ra :
$m_{hỗn\ hợp}=m_{Al}+m_{Al_2O_3}=0,15.27+0,1.102=14,25(gam)$
Vậy :
$\%m_{Al} = \dfrac{0,15.27}{14,25}.100\% = 28,42\%$
$\%m_{Al_2O_3} = 100\% - 28,42\% = 71,58\%$