Câu $\rm 1$ 

$\rm n_A=\dfrac{5,4}{M_A}(mol)$ 

$\rm n_{H_2}=\dfrac{0,6}{2}=0,3(mol)$ 

Phương trình hóa học 

$\rm 2A +3H_2SO_4 \to A_2(SO_4)_3 +3H_2$ 

Theo phương trình hóa học 

$\rm n_{A}=\dfrac{2}{3}n_{H_2}$ 

`<=>`$\rm \dfrac{5,4}{M_A}=0,2$ 

`->`$\rm M_A=27(Al)$ 

`1)` Kim loại cần tím là $\rm Al$ (Aluminium) 

`2)` Theo phương trình hóa học 

$\rm n_{H_2SO_4}=n_{H_2}=0,3(mol)$ 

`->`$\rm m_{dung~dịch~H_2SO_4}=\dfrac{0,3.98}{12\%}=245(g)$ 

`3)` 

$\rm m_{dung~dịch~sau~phản~ứng}=5,4+245-0,6=249,8(g)$ 

$\rm m_{Al_2(SO_4)_3}=0,1.342=34,2(g)$ 

`->`$\rm C\%_{Al_2(SO_4)_3}=\dfrac{34,2}{249,8}.100\%=13,69\%$ 




Câu $\rm 2$ 

$\rm (1) 4Al +3O_2 \xrightarrow{to} 2Al_2O_3$ 

$\rm (2)Al_2O_3 \xrightarrow[Na_3AlF_4]{đpnc} 2Al +\dfrac{3}{2}O_2$ 

$\rm (3)Al_2O_3 +6HCl \to 2AlCl_3 +3H_2O$ 

$\rm (4)Al_2O_3 +3H_2SO_4 \to Al_2(SO_4)_3 +3H_2O$ 

$\rm (5) 2Al +3H_2SO_4 \to Al_2(SO_4)_3 +3H_2$ 

$\rm (6) Al_2(SO_4)_3 +3BaCl_2 \to 2AlCl_3 +3BaSO_4$ 

$\rm (7) AlCl_3 + 3NaOH \to Al(OH)_3 +3NaCl$ 

$\rm (8) Al(OH)_3 +3HCl \to AlCl_3 +3H_2O $ 

$\rm (9)Al_2(SO_4)_3 +3Ba(OH)_2 \to 2Al(OH)_3 +3BaSO_4$ 

$\rm (10)2Al(OH)_3 +3H_2SO_4 \to Al_2(SO_4)_3 +3H_2o$ 

$\rm (11) 2Al(OH)_3 \xrightarrow{to} Al_2O_3 +3H_2O$ 

$\rm (12) Al_2O_3 + 3HNO_3 \to Al(NO_3)_3 +3H_2O$ 

$\rm (13) Al(OH)_3 +3HNO_3 \to Al(NO_3)_3 +3H_2O$ 

$\rm (14)Al(NO_3)_3 +3NaOH \to Al(OH)_3 +2NaNO_3$ 

$\rm (15) Al(NO_3)_3 +3AgCl \to AlCl_3 +3AgNO_3$ 

$\rm (16) 2Al +3S \xrightarrow{to} Al_2S_3$ 

$\rm (17) 2Al +6HCl \to 2AlCl_3 +3H_2$ 

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