Đáp án:

 a=91,25 gam

\({{\text{V}}_{Ba{{(OH)}_2}}}{\text{ = 125ml}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(MgO + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}O\)

Ta có: \({n_{MgO}} = \frac{{10}}{{24 + 16}} = 0,25{\text{ mol}} \to {{\text{n}}_{HCl}} = 2{n_{MgO}} = 0,5{\text{ mol}}\)

\( \to {m_{HCl}} = 0,5.36,5 = 18,25{\text{ gam}} \to {{\text{m}}_{dd{\text{ HCl}}}} = \frac{{18,25}}{{20\% }} = 91,25{\text{ gam = a}}\)

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

\( \to {n_{Ba{{(OH)}_2}}} = \frac{1}{2}{n_{HCl}} = 0,25{\text{ mol}} \to {{\text{V}}_{Ba{{(OH)}_2}}} = \frac{{0,25}}{2} = 0,125{\text{ lít= 125ml}}\)