Đáp án:
\(\begin{array}{l}
a)\\
{m_{BaO}} = 15,3g\\
b)\\
C{\% _{Ba{{(OH)}_2}}} = 3,42\% \\
C{\% _{BaC{l_2}}} = 4,16\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
BaO + {H_2}O \to Ba{(OH)_2}\\
{m_{Ba{{(OH)}_2}}} = \frac{{8,55 \times 200}}{{100}} = 17,1g\\
{n_{Ba{{(OH)}_2}}} = \frac{m}{M} = \frac{{17,1}}{{171}} = 0,1mol\\
\Rightarrow {n_{BaO}} = {n_{Ba{{(OH)}_2}}} = 0,1mol\\
{m_{BaO}} = n \times M = 0,1 \times 153 = 15,3g\\
b)\\
Ba{(OH)_2} + 2HCl \to BaC{l_2} + 2{H_2}O\\
{m_{HCl}} = \dfrac{{50 \times 7,3}}{{100}} = 3,65g\\
{n_{HCl}} = \dfrac{m}{M} = \dfrac{{3,65}}{{36,5}} = 0,1mol\\
\dfrac{{0,1}}{1} > \dfrac{{0,1}}{2} \Rightarrow Ba{(OH)_2}\text{ dư}\\
{n_{Ba{{(OH)}_d}}} = {n_{Ba{{(OH)}_2}}} - \dfrac{{{n_{HCl}}}}{2} = 0,05mol\\
{m_{Ba{{(OH)}_2}}} = n \times M = 0,05 \times 171 = 8,55g\\
{n_{BaC{l_2}}} = \dfrac{{{n_{HCl}}}}{2} = 0,05mol\\
{m_{BaC{l_2}}} = n \times M = 0,05 \times 208 = 10,4g\\
{m_{{\rm{dd}}spu}} = 200 + 50 = 250g\\
C{\% _{Ba{{(OH)}_2}}} = \dfrac{m}{{{m_{{\rm{dd}}spu}}}} \times 100\% = \dfrac{{8,55}}{{250}} \times 100\% = 3,42\% \\
C{\% _{BaC{l_2}}} = \dfrac{m}{{{m_{{\rm{dd}}spu}}}} \times 100\% = \dfrac{{10,4}}{{250}} \times 100\% = 4,16\%
\end{array}\)
