\(\begin{array}{l}
1)\\
a)\\
MgC{l_2} + 2NaOH \to Mg{(OH)_2} + 2NaCl\\
nNaOH = \frac{{200 \times 10\% }}{{40}} = 0,5\,mol\\
nMgC{l_2} = \frac{{0,5}}{2} = 0,25\,mol\\
m{\rm{dd}}MgC{l_2} = \frac{{0,25 \times 95}}{{4\% }} = 593,75g\\
b)\\
nMg{(OH)_2} = nMgC{l_2} = 0,25\,mol\\
mMg{(OH)_2} = 0,25 \times 58 = 14,5\,g\\
c)\\
m{\rm{dd}}spu = 200 + 593,75 - 14,5 = 779,25g\\
C\% NaCl = \frac{{0,5 \times 58,5}}{{779,25}} \times 100\%  = 3,75\% \\
2)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
a)\\
nMg = \frac{{2,4}}{{24}} = 0,1\,mol\\
n{H_2} = 0,1\,mol\\
V{H_2} = 0,1 \times 22,4 = 2,24l\\
b)\\
m{\rm{dd}}HCl = \frac{{0,2 \times 36,5}}{{20\% }} = 36,5g\\
c)\\
m{\rm{dd}}spu = 36,5 + 2,4 - 0,1 \times 2 = 38,7g\\
C\% MgC{l_2} = \frac{{0,1 \times 95}}{{38,7}} \times 100\%  = 24,55\% 
\end{array}\)