Đáp án:

a)

\(\begin{array}{l}
{m_{Mg{{\left( {OH} \right)}_2}}} = 1,16g\\
{m_{MgO}} = 0,8g\\
b)\\
{C_{{M_{NaCl}}}} = 0,2M\\
{C_{{M_{NaOH}}}} = 3,55M
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
MgC{l_2} + 2NaOH \to Mg{(OH)_2} + 2NaCl\\
{n_{MgC{l_2}}} = {C_M} \times V = 0,1 \times 0,2 = 0,02mol\\
{n_{NaOH}} = \dfrac{m}{M} = \dfrac{{30}}{{40}} = 0,75mol\\
\dfrac{{0,02}}{1} < \dfrac{{0,75}}{2} \Rightarrow NaOH\text{ dư}\\
 \Rightarrow {n_{Mg{{(OH)}_2}}} = {n_{MgC{l_2}}} = 0,02mol\\
{m_{Mg{{\left( {OH} \right)}_2}}} = n \times M = 0,02 \times 58 = 1,16g\\
Mg{(OH)_2} \to MgO + {H_2}O\\
{n_{MgO}} = {n_{Mg{{(OH)}_2}}} = 0,02mol\\
{m_{MgO}} = n \times M = 0,02 \times 40 = 0,8g\\
b)\\
{n_{NaCl}} = 2{n_{MgC{l_2}}} = 2 \times 0,02 = 0,04mol\\
{C_{{M_{NaCl}}}} = \dfrac{n}{V} = \dfrac{{0,04}}{{0,2}} = 0,2M\\
{n_{NaOHpu}} = 2{n_{MgC{l_2}}} = 2 \times 0,02 = 0,04mol\\
{n_{NaO{H_d}}} = {n_{NaOH}} - {n_{NaOHpu}} = 0,75 - 0,04 = 0,71mol\\
{C_{{M_{NaOH}}}} = \dfrac{n}{V} = \dfrac{{0,71}}{{0,2}} = 3,55M
\end{array}\)