Đặt `n_{H_2SO_4}=x(mol); n_{SO_4^{2-}((Y)}=y(mol)`

Bảo toàn nguyên tố `S:` `n_S=(0,25+y-x)(mol)`

`->m_{KL(X)}=5,21-32.(0,25+y-x)=(32x-32y-2,79)(g)`

BTNT S: `n_{BaSO_4}=n_{SO_4^{2-}(Y)}=y(mol)`

BTNT Ba: `n_{Ba(OH)_2}=n_{BaSO_4}=y(mol)`

`->n_{OH^{-}}=2y(mol)`

Ta có: `m_{\downarrow}=m_{BaSO_4}+m_{KL}+m_{OH^{-}}`

`->21,98=233y+32x-32y-2,79+17.2y`

`->32x+235y=24,77(1)`

BTNT H: `n_{H_2O}=n_{H_2SO_4}=x(mol)`

BTKL: `5,21+98x=0,25.64+18x+96y+32x-32y-2,79`

`->48x-64y=8(2)`

`(1),(2)->x=0,26; y=0,07`

`m=0,07.96+32.0,26-32.0,07-2,79=10,01(g)`