`n_{H_2S}={8,98}/{22,4}=0,4(mol)`
`2H_2S+3O_2->2SO_2+2H_2O`
`n_{SO_2}=n_{H_2S}=0,4(mol)`
`n_{NaOH}={80.1,28.25\%}/{40}=0,625(mol)`
`T={0,625}/{0,4}=1,5625`
`->` Tạo `Na_2SO_3:x(mol);NaHSO_3:y(mol)`
BT Na: `2x+y=0,625(1)`
BT S: `x+y=0,4(2)`
`(1)(2)->x=0,225(mol);y=0,175(mol)`
`m_{dd\ spu}=0,4.64+80.1,28=128(g)`
`C\%_{Na_2SO_3}={0,225.126}/{128}.100\%\approx 22,15\%`
`C\%_{NaHSO_3}={0,175.104}/{128}.100\%\approx 14,22\%`