Đáp án:

\({C_{M{\text{ Ba(OH}}{{\text{)}}_2}}} = 0,333M\)

\({m_{BaC{l_2}}} = 10,4{\text{ gam}}\)

\({C_{M{\text{ BaC}}{{\text{l}}_2}}}= 0,25M\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(Ba{(OH)_2} + 2HCl\xrightarrow{{}}BaC{l_2} + 2{H_2}O\)

Ta có:

\({n_{HCl}} = 0,05.2 = 0,1{\text{ mol}}\)

\( \to {n_{Ba{{(OH)}_2}}} = {n_{BaC{l_2}}} = \frac{1}{2}{n_{HCl}} = 0,05{\text{ mol}}\)

\( \to {C_{M{\text{ Ba(OH}}{{\text{)}}_2}}} = \frac{{0,05}}{{0,15}} = 0,333M\)

\( \to {m_{BaC{l_2}}} = 0,05.(137 + 35,5.2) = 10,4{\text{ gam}}\)

\({V_{dd}} = 150 + 50 = 200{\text{ ml = 0,2 lít}}\)

\( \to {C_{M{\text{ BaC}}{{\text{l}}_2}}} = \frac{{0,05}}{{0,2}} = 0,25M\)

$\text{n$_{HCl}$=0,05.2=0,1(mol)}$

$\text{a,PTHH:}$

$\text{Ba(OH)$_{2}$+2HCl→BaCl$_{2}$+2H$_{2}$O}$

$\text{0,05←  0,1  →0,05  (mol)}$

$\text{b,C$_{M Ba(OH)2}$=$\frac{0,05}{0,15}$= $\frac{1}{3}$ (M)}$

$\text{c,m$_{BaCl2}$=0,05.208=10,4(g)}$

$\text{d,C$_{M BaCl2}$=$\frac{0,05}{0,15+0,05}$= 0,25(M)}$