$PTHH : 2Al +3H_2SO_4 \xrightarrow{}Al_2(SO_4)_3+3H_2(1)$

$Al_2O_3 +3H_2SO_4 \xrightarrow{} Al_2(SO_4)_3 +3H_2O(2)$

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$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$(sửa đề $6,72(lít)$)

$a) n_{Al}=\dfrac{2}{3}n_{H_2}=0,2(mol)$ 

$m_{Al}=0,2 ×27 =5,4(g)$ 

$n_{H_2SO_4}=0,4×3=1,2(mol)$ 

$n_{H_2SO_4(1)}=n_{H_2}=0,3(mol)$ 

$⇒n_{H_2SO_4(2)}=1,2-0,3=0,9(mol)$ 

$→n_{Al_2O_3}=\dfrac{1}{3}n_{H_2SO_4(2)}=0,3(mol)$ 

$m_{Al_2O_3}=0,3×102=30,6(g)$ 

$b)$ta có $n_{Al_2(SO_4)_3}=n_{H_2SO_4}=1,2(mol)$ 

$C_{M_{Al_2(SO_4)_3}}=\dfrac{1,2}{0,4}=3M$

`a)n_{H_{2}}=\frac{6,72}{22,4}=0,3(mol)`

`n_{H_{2}SO_{4}}=0,4.3=1,2(mol)`

`Al_{2}O_{3}+3H_{2}SO_{4}→Al_{2}(SO_{4})_{3}+3H_{2}O(1)`

`2Al+3H_{2}SO_{4}→Al_{2}(SO_{4})_{3}+3H_{2}(2)`

`0,2`                                                              `0,3`    / `mol`

`⇒m_{Al}=0,2.27=5,4(g)`

Theo phương trình `(2)⇒n_{H_{2}SO_{4}(2)}=n_{H_{2}}=0,3(mol)`

`⇒n_{H_{2}SO_{4}(1)}=1,2-0,3=0,9(mol)`

Theo phương trình `(2)⇒n_{Al_{2}O_{3}}=\frac{1}{3}.n_{H_{2}SO_{4}(1)}=0,6(mol)`

`⇒m_{Al_{2}O_{3}}=0,6.102=61,2(g)`

`b)` Theo phương trình `(1);(2)⇒n_{Al_{2}(SO_{4})_{3}}=\frac{1}{3}.n_{H_{2}SO_{4}}=0,4(mol)`

`⇒C_{M_{ddAl_{2}(SO_{4})_{3}}}=\frac{0,4}{0,4}=1(M)`