`#AkaShi`

PTHH:

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

`->m_{HCl}=m_{dd}xxC%=200xx14,6%=29,2\ (g)`

`->n_{HCl}=(m_{HCl})/(M_{HCl})=(29,2)/(36,5)=0,8\ (mol)`

`->m_{Ba(OH)_2}=m_{dd}xxC%=200xx17,1%=34,2\ (g)`

`->n_{Ba(OH)_2}=(m_{Ba(OH)_2})/(M_{Ba(OH)_2})=(34,2)/(171)=0,2\ (mol)`

Xét hết dư:

`->n_{HCl}=(0,8)/2=0,4 > n_{Ba(OH)_2}=0,2\ (mol)`

`->HCl` dư và `Ba(OH)_2` hết

Sau phản ứng gồm dung dịch:

`{(BaCl_2),(HCl\ dư):}`

`->m_{dd}=m_{dd\ HCl}+m_{dd\ Ba(OH)_2}=200+200=400\ (g)`

`->n_{BaCl_2}=n_{Ba(OH)_2}=0,2\ (mol)`

`->n_{HCl\ dư}=2xxn_{Ba(OH)_2}=0,2xx2=0,4\ (mol)`

`->n_{HCl\ dư}=0,8-0,4=0,4\ (mol)`

`->C%_{HCl\ dư}=(0,4xx36,5)/400xx100%=3,65%`

`->C%_{BaCl_2}=(0,2xx208)/400xx100%=10,4%`

Đáp án:

\(\begin{array}{l}
{C_\% }HCl = 3,65\% \\
{C_\% }BaC{l_2} = 10,4\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
Ba{(OH)_2} + 2HCl \to BaC{l_2} + 2{H_2}O\\
{n_{Ba{{(OH)}_2}}} = \dfrac{{200 \times 17,1\% }}{{171}} = 0,2\,mol\\
{n_{HCl}} = \dfrac{{200 \times 14,6\% }}{{36,5}} = 0,8\,mol\\
{n_{Ba{{(OH)}_2}}} < \dfrac{{{n_{HCl}}}}{2} \Rightarrow HCl \text{ dư }\\
{n_{HCl}} \text{ dư }= 0,8 - 0,2 \times 2 = 0,4\,mol\\
{n_{BaC{l_2}}} = {n_{Ba{{(OH)}_2}}} = 0,2\,mol\\
{m_{{\rm{dd}}spu}} = 200 + 200 = 400g\\
{C_\% }HCl \text{ dư }= \dfrac{{0,4 \times 36,5}}{{400}} \times 100\%  = 3,65\% \\
{C_\% }BaC{l_2} = \dfrac{{0,2 \times 208}}{{400}} \times 100\%  = 10,4\% 
\end{array}\)