Giải giúp tuiiiiiiii

$a)$

$\text{ x  2 |  }$$\mathop{Mn}\limits^{+7}++5e \to \mathop{Mn}\limits^{+2} $

$\text{ x  5 |  }$$\mathop{S}\limits^{+4} \to \mathop{S}\limits^{+6}+2e $

`->`$5SO_2+2KMnO_4+2H_2O \to 2H_2SO_4+K_2SO_4+2MnSO_4$

$b)$

Bảo toàn e:

$2n_{SO_2}=5n_{KMnO_4}$

`=>`$n_{SO_2}=\dfrac{5 \times 0.1 \times 0.02 }{2}=0.005(mol)$

`=>`$V_{SO_2}=0.005 \times 24.79=0.12395(L)$