làm giúp em bài 14 vs ạ

`500mL=0,5L`

`n_{Ba(OH)_2}=0,5.0,1=0,05(mol)`

TH1: `Al_2(SO_4)_3` dư.

`3Ba(OH)_2+Al_2(SO_4)_3->3BaSO_4+2Al(OH)_3`

Theo PT: `n_{BaSO_4}=0,05(mol);n_{Al(OH)_3}=1/{30}(mol)`

`->m_↓=0,05.233+1/{30}.78=14,25>12,045`

Loại.

TH2: `Ba(OH)_2` dư.

Đặt `n_{Al_2(SO_4)_3}=x(mol)`

`3Ba(OH)_2+Al_2(SO_4)_3->3BaSO_4+2Al(OH)_3(1)`

`3x←x->3x->2x(mol)`

`2Al(OH)_3+Ba(OH)_2->Ba(AlO_2)_2+4H_2O(2)`

`->m_{BaSO_4}=3x.233=699x(g)`

`->n_{Al(OH)_3\ spu}={12,045-699x}/{78}(mol)`

`->n_{Al(OH)_3(2)}=2x-{12,045-699x}/{78}(mol)`

Lại có `\sum n_{Ba(OH)_2}=0,5n_{Al(OH)_3(2)}+3x`

`->0,05=x-{12,045-699x}/{156}+3x`

`->x=0,015(mol)`

`->V_{dd\ Al_2(SO_4)_3}={0,015}/{0,1}=0,15(L)=150(mL)`

Vậy `V=150`