Đáp án:

\(\begin{array}{l}
a)\\
\% {m_{Mg}} = 21,32\% \\
\% {m_{BaC{O_3}}} = 78,68\% \\
b)\\
{m_{Mg{{(OH)}_2}}} = 8,7g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
BaC{O_3} + 2HCl \to BaC{l_2} + C{O_2} + {H_2}O\\
{n_{HCl}} = 0,4 \times 1,5 = 0,6\,mol\\
hh:Mg(a\,mol),BaC{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
24a + 197b = 23,3\\
2a + 2b = 0,6
\end{array} \right.\\
 \Rightarrow a = 0,207;b = 0,093\\
\% {m_{Mg}} = \dfrac{{0,207 \times 24}}{{23,3}} \times 100\%  = 21,32\% \\
\% {m_{BaC{O_3}}} = 100 - 21,32 = 78,68\% \\
b)\\
MgC{l_2} + 2NaOH \to Mg{(OH)_2} + 2NaCl\\
{n_{NaOH}} = 0,3 \times 1 = 0,3\,mol\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,207\,mol\\
{n_{MgC{l_2}}} > \dfrac{{{n_{NaOH}}}}{2} \Rightarrow MgC{l_2} \text{ dư }\\
{n_{Mg{{(OH)}_2}}} = \dfrac{{0,3}}{2} = 0,15\,mol\\
{m_{Mg{{(OH)}_2}}} = 0,15 \times 58 = 8,7g
\end{array}\)