Đáp án:

 `a,3,2(lít)`

`b,208,8(g)`

`c,0,22M`

Giải thích các bước giải:

 `n_{Ba(OH)_2}=(400)/(1000).2=0,8(mol)`

PTHH:
`Ba(OH)_2+2HNO_3->Ba(NO_3)_2+2H_2O`

`0,8`        `->1,6`       `->0,8(mol)`

`a,`

`V_{ddHNO_3(0,5M)}=(1,6)/(0,5)=3,2(lít)`

`b,`

`m_{m uối}=m_{Ba(NO_3)_2}=0,8.261=208,8(g)`

`c,`

`V_{dd(sau)}=0,4+3,2=3,6(lít)`

`C_{M Ba(NO_3)_2}=(0,8)/(3,6)≈0,22(M)`

`#Ly`

$a)$

Đổi $400$ $ml$ $=$ $0,4$ $l$

`n_(Ba(OH)_2` `=` `0,4.2=0,8` $(mol)$

PTHH: $Ba(OH)_2$ $+$ $2HNO_3$ `->` $Ba(NO_3)_2$ $+$ $2H_2O$

`->` `n_(HNO_3` `=` `2n_(Ba(OH)_2` `=` `1,6` $(mol)$

`V_(ddHNO_3` ``= `(1,6)/(0,5)=3,2` $(l)$

$b)$

Theo PT, `n_(Ba(NO_3)_2` `=` `n_(Ba(OH)_2` `=` `0,8` $(mol)$

`m_(Ba(NO_3)_2` `=` `0,8.261=208,8` $(g)$

$c)$

`V_(ddBa(NO_3)_2` `=` `0,4+3,2=3,6` $(l)$

`C_MBa(NO_3)_2` ``= `(0,8)/(3,6)=0,22` $(M)$