`b)`

`n_{SO_2}={2,9748}/{24,79}=0,12(mol)`

Bte: `3n_{Al}=2n_{SO_2}=0,24(mol)`

`->n_{Al}=0,08(mol)`

`->n_{Al_2O_3}={6,24-0,08.27}/{102}=0,04(mol)`

BTNT Al: `n_{Al_2(SO_4)_3}=0,04+0,08.0,5=0,08(mol)`

BTNT S: `n_{H_2SO_4}=3n_{Al_2(SO_4)_3}+n_{SO_2}=0,36(mol)`

`m_{dd\ sau}=6,24+{0,36.98}/{80\%}-0,12.64=42,66(g)`

`C\%_{Al_2(SO_4)_3}={0,08.342}/{42,66}.100\%=64,135\%`

$10.$

$a.$

Bảo toàn $e:$

$n_{Al}=\dfrac{2}{3}n_{SO_2}=0.08(mol)$

$n_{Al_2O_3}=\dfrac{6.24-0.08 \times 27}{102}=0.04(mol)$

$m_{Al}=2.16(gam);m_{Al_2O_3}=4.08(gam)$

$b.$

$n_{H_2SO_4,oxh-k}=2n_{SO_2}=0.24(mol)$

$n_{H_2SO_4,acid-baz}=3n_{Al_2O_3}=0.12(mol)$

`=>`$m_{dd,H_2SO_4}=\dfrac{(0.24+0.12) \times 98 \times 100}{80}=44.1(gam)$

Bảo toàn khối lượng:

$m_{Al}+m_{Al_2O_3}+m_{dd,H_2SO_4}=m_{dd}+m_{SO_2}$

`=>`$m_{dd}=42.66(gam)$

Bảo toàn nguyên tố $Al:$

$n_{Al_2(SO_4)_3}=\dfrac{n_{Al}+2n_{Al_2O_3}}{2}=0.08(mol)$

`=>`$C\%(Al_2(SO_4)_3)=64.14\%$

$c.$

$n_{KOH}=\dfrac{70 \times 12\%}{56}=0.15(mol)$

$T=\dfrac{0.12}{0.15}<1$

`=>` Chỉ có muối $HSO_3^-$

Bảo toàn nguyên tố $S:$

$n_{KHSO_3}=n_{SO_2}=0,12(mol)$

Bảo toàn khối lượng:

$m_{SO_2}+m_{dd,KOH}=m_{dd}$

`=>`$m_{dd}=77.68(gam)$

`=>`$C\%(KHSO_3)=18.54\%$