`a)` 

`H_2 + Cl_2→ 2HCl` 

 `A: H_2(x), Cl_2(y)`

Theo pt: 

`n_{H_2}=  n_{Cl_2}= z(mol)`

`n_A = {6,72}/{22,4} = 0,3(mol)`

`n_A  = x + y= 0,3(mol) (1)`

`n_{Cl_2} (B) = n_{Cl_2} (A) - n_{H_2}` 

     `= y - z = 0,4y(2)`

`n_B = n_A = x + y = {6,72}/{22,4}(mol)`

Hay 

`x + y - 8z = 0 (3)`

 `(1)(2)(3)`

$→\begin{cases} x = 0,2375(mol)\\y = 0,0625(mol)\\z = 0,0375(mol) \end{cases}$

$→ \begin{cases} n_{H_2} (A) = 0,2375(mol)\\n_{Cl_2} (A) = 0,0625(mol)\\n_{HCl} =   0,075(mol) \\n_{Cl_2} (B) = 0,025(mol )\end{cases}$

`%V_{Cl_2} (A) = {0,0625}/{0,3} .100= 20,83%`

`%V_{H_2} (A) = 100 - 20,83% = 79,17%`

`%V_{Cl_2} (B) = {0,025}/{0,3} .100= 8,33%`

`%V_{HCl} = {0,075}/{0,3} .100 = 25%`

`%V_{H_2} = 100 -25-8,33=66,67%`

`b)` 

`2NaOH + Cl_2 → NaCl + NaClO + H_2O (1)`

`HCl + NaOH → NaCl + H_2O (2)`

`n_{NaOH} = {20% . 30}/{40} = 0,15(mol)` 

`(2)`: 

`n_{NaOH} > n_{HCl}`

`→ NaOH` dư 

`n_{NaCl}  = n_{HCl} =  0,075(mol)`

`n_{NaOH} (du)= 0,15 -  0,075= 0,075(mol)`

`(1)`: 

`n_{Cl_2}  < n_{NaOH}`

`→ NaOH`  dư 

`n_{NaCl} = n_{NaClO} =n_{Cl_2} =  0,025(mol)`

`n_{NaOH}(du) = 0,075 - 0,025 .2 = 0,025( mol)` 

`mdd = 0,075 . 36,5 + 0,025 . 71 + 30`

        `= 34,5125(g)`

`C%_{NaCl} = {(0,025 + 0,075) .58,5}/{34,5125} . 100  = 16,95%`

`C%_{NaClO} = {0,025 . 74,5}/{34,5125}. 100 = 5,4%`

`C%_{NaOH}= {0,025. 40}/{34,5125} . 100  =2,9%`