`a)`

`2R + 2nHCl → 2RCl_n + nH_2`

`n_{H_2} = {13,44}/{22,4} = 0,6(mol)`

`m↑= m_R - m_{H_2}`

`<=> 9,6 = m - 0,6.  2`

`→ m = 10,8(g)`

Theo pt: 

`n_R = {2n_{H_2}}/n `

`= {0,6 . 2}/n = {1,2}/n(mol)`

`M_R = {10,8}/{{1,2}/n} = 9n`$(g/mol)$

`n = 3 → M_R = 27`$(g/mol)$

`→ R` là nhôm `(Al)`

Bảo toàn `H`

`n_{HCl} = 2n_{H_2} = 0,6 . 2 = 1,2(mol)`

`V_{HCl}  = {1,2}/{0,5} = 2,4(l)`

`b)`

Bảo toàn `Al`

`n_{AlCl_3} = n_{Al}`

`= {10,8}/{27} = 0,4(mol)`

`n_{text{muối tách}} = 25% . 0,4 = 0,1(mol)`

Muối tách ra dạng `AlCl_3 . nH_2O`

`M_T = {24,15}/{0,1} = 241,5`$(g/mol)$

`M_T = 18n + 133,5 = 241,5`$(g/mol)$

`→ n = 6`

Muối là `AlCl_3 . 6H_2O`

`Al_2(SO_4)_3 + 6NaOH → 2Al(OH)_3 + 3Na_2SO_4(1)`

`n_{Al_2(SO_4)_3} = {100}/{1000} . 1 = 0,1(mol)`

`n_{Al(OH)_3} = {12,48}/{78} = 0,16(mol)`

`n_{Al_2(SO_4)_3} ≠ 1/2n_{Al(OH)_3}`

`→` Có hai trường hợp xảy ra

Trường hợp 1: `Al_2(SO_4)_3` dư

Theo pt: 

`n_{NaOH} = 3n_{Al(OH)_3}`

`= 0,16 . 3 = 0,48(mol)`

`x = {0,48}/{{200}/{1000}} = 2,4(M)`

Trường hợp 2: `NaOH` dư 

Xảy ra `(1)` 

`Al(OH)_3 + NaOH → NaAlO_2 + 2H_2O`

Bảo toàn `Al`

`n_{Al(OH)_3} = 2n_{Al_2(SO_4)_3}`

`= 0,1 .2 = 0,2(mol)`

`n_{Al(OH)_3} (text{tan}) = 0,2 - 0,16 = 0,04(mol)`

Theo pt: 

`n_{NaOH} = 6n_{Al_2(SO_4)_3} + n_{Al(OH)_3} (text{tan})`

`= 0,1 . 6  + 0,04 = 0,64(mol)`

`x= {0,64}/{{200}/{1000}} =3,2(M)`

\(\begin{array}{l}
4)\\
a)\\
n_{H_2}=\frac{13,44}{22,4}=0,6(mol)\\
m_{dd\,tang}=m_R-m_{H_2}\\
\to m=m_R=9,6+0,6.2=10,8(g)\\
Dat\,hoa\,tri\,R\,la\,n\\
2R+2nHCl\to 2RCl_n+nH_2\\
Theo\,PT:\,n_R=\frac{2n_{H_2}}{n}=\frac{1,2}{n}\\
\to M_R=9n\to n=3;M_R=27\\
\to R:Al\\
Theo\,PT:\,n_{HCl}=2n_{H_2}=1,2(mol)\\
\to V_{dd\,HCl}=\frac{1,2}{0,5}=2,4(l)\\
b)\\
Theo\,PT:\,n_{AlCl_3}=\frac{2}{3}n_{H_2}=0,4(mol)\\
\to n_{H_2O(T)}=\frac{24,15-0,4.133,5.25\%}{18}=0,6(mol)\\
\to n_{AlCl_3(T)}:n_{H_2O(T)}=(0,4.25\%):0,6=1:6\\
\to T:AlCl_3.6H_2O\\
5)\\
200ml=0,2l;100ml=0,1l\\
n_{Al_2(SO_4)_3}=0,1.1=0,1(mol)\\
n_{Al(OH)_3}=\frac{12,48}{78}=0,16(mol)\\
Al_2(SO_4)_3+6NaOH\to 2Al(OH)_3+3Na_2SO_4\\
TH_1:\rm Chỉ\,tạo\,ktua\\
\to n_{NaOH}=3n_{Al(OH)_3}=0,48(mol)\\
\to C_{M\,NaOH}=\frac{0,48}{0,2}=2,4M\\
TH_2:\rm Kết tủa bị hòa tan\\
Al_2(SO_4)_3+6NaOH\to 2Al(OH)_3+3Na_2SO_4\\
Al(OH)_3+NaOH\to NaAlO_2+H_2O\\
Bao\,toan\,Al:\,n_{NaAlO_2}=2n_{Al_2(SO_4)_3}-n_{Al(OH)_3}=0,04(mol)\\
Theo\,PT:\,n_{NaOH}=3n_{Al(OH)_3}+4n_{NaAlO_2}=0,64(mol)\\
\to C_{M\,NaOH}=\frac{0,64}{0,2}=3,2M
\end{array}\)