Đáp án:

$\downarrow$

Giải thích các bước giải:

* `n_{HNO_3}=189/63=3(mol)`

* `n_{KOH}=112/56=2(mol)`

`-` PTHH:    `HNO_3+KOH->KNO_3+H_2O`

`@` Tỷ lệ:   `3/1 > 2/1` `->` `HNO_3`  dư

`->n_{HNO_3(p.ư)}=n_{KOH}=2(mol)`

`->n_{HNO_2(dư)}=3-2=1(mol)`

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

`->n_{Ba(OH)_2}=1/2 n_{HNO_3(dư)}=0,5(mol)`

`->m_{Ba(OH)_2}=0,5xx171=85,5(gam)`

`->m_{dd Ba(OH)_2}=(85,5)/(25%) = 342(gam)`

$\text{#Kudo}$

Em tham khảo nha :

\(\begin{array}{l}
KOH + HN{O_3} \to KN{O_3} + {H_2}O\\
{n_{KOH}} = \dfrac{{112}}{{56}} = 2mol\\
{n_{HN{O_3}}} = \dfrac{{189}}{{63}} = 3mol\\
{n_{HN{O_3}d}} = 3 - 2 = 1mol\\
Ba{(OH)_2} + 2HN{O_3} \to Ba{(N{O_3})_2} + 2{H_2}O\\
{n_{Ba{{(OH)}_2}}} = \dfrac{{{n_{HN{O_3}d}}}}{2} = 0,5mol\\
{m_{Ba{{(OH)}_2}}} = 0,5 \times 171 = 85,5g\\
{m_{{\rm{dd}}Ba{{(OH)}_2}}} = \dfrac{{85,5 \times 100}}{{25}} = 342g
\end{array}\)