Đáp án + Giải thích các bước giải:

$a\bigg)$

$n_{KOH}=\frac{560.20\%}{56}=2(mol)$

$\sum n_{HCl}=\frac{365.30\%}{36,5}=3(mol)$

$KOH+HCl\to KCl+H_2O(1)$

$Ba(OH)_2+2HCl\to BaCl_2+2H_2O(2)$

Theo (1): $n_{HCl(1)}=n_{KOH}=2(mol)$

$\to n_{HCl(2)}=3-2=1(mol)$

Theo (2): $n_{Ba(OH)_2}=\frac{1}{2}n_{HCl}=0,5(mol)$

$\to m_{dd\,Ba(OH)_2}=\frac{0,5.171}{25\%}=342(g)$

$b\bigg)$

Theo PT: $n_{KCl}=n_{KOH}=2(mol);n_{BaCl_2}=n_{Ba(OH)_2}=0,5(mol)$

$\to C\%_{KCl}=\frac{2.74,5}{560+365}.100\%\approx 16,11\%$

$\to C\%_{BaCl_2}=\frac{0,5.208}{342+365}.100\%\approx 14,71\%$

Đáp án:

 `↓`

Giải thích các bước giải:

 `m_{HCl}=365.30\%=109,5(g)`

`->n_{HCl}=(109,5)/(36,5)=3(mol)`

`m_{KOH}=560.20\%=112(g)`

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

PTHH:

`KOH+HCl->KCl+H_2O`

`2->2->2(mol)`

`->n_{HCl sau pứ}=3-2=1(mol)`

`Ba(OH)_2+2HCl->BaCl_2+2H_2O`

`0,5←1->0,5(mol)`

a,Từ PTHH ta suy ra :

`n_{Ba(OH)_2}=0,5(mol)`

`->m_{Ba(OH)_2}=0,5.171=85,5(g)`

`->m_{ddBa(OH)_2}=85,5:25\%=342(g)`

b,Từ PTHH :

`n_{KCl}=2(mol)->m_{KCl}=2.74,5=149(g)`

`n_{BaCl_2}=0,5(mol)->m_{BaCl_2}=0,5.208=104(g)`

Ta có : `m_{dd sau pứ}=365+560+342=1267(g)`

`⇒C\%_{KCl}=(149)/(1267).100\%≈11,76\%`

`C\%_{BaCl_2}=(104)/(1267).100\%≈8,21\%`