\(\begin{array}{l}
18)\\
Mg + 2HCl \to MgC{l_2} + {H_2}(1)\\
MgO + 2HCl \to MgC{l_2} + {H_2}O(2)\\
n{H_2} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
nMgC{l_2} = \dfrac{{28,5}}{{95}} = 0,3\,mol\\
nMgC{l_2}(1) = n{H_2} = 0,1\,mol\\
nMgC{l_2}(2) = 0,3 - 0,1 = 0,2\,mol\\
nMg = nMgC{l_2}(1) = 0,1\,mol\\
nMgO = nMgC{l_2}(2) = 0,2\,mol\\
a = 0,1 \times 24 + 0,2 \times 40 = 10,4g\\
nHCl = 0,1 \times 2 + 0,2 \times 2 = 0,6\,mol\\
x = {C_M}HCl = \dfrac{{0,6}}{{0,4}} = 1,5M\\
19)\\
a)\\
2Na + 2{H_2}O \to 2NaOH + {H_2}(1)\\
N{a_2}O + {H_2}O \to 2NaOH(2)\\
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O\\
nN{a_2}C{O_3} = \dfrac{{6,36}}{{106}} = 0,06\,mol\\
 \Rightarrow nNaOH = 2 \times 0,06 = 0,12\,mol\\
n{H_2} = \dfrac{{0,336}}{{22,4}} = 0,015\,mol \Rightarrow nNaOH(1) = 0,03\,mol\\
nNaOH(2) = 0,12 - 0,03 = 0,09\,mol\\
nNa = nNaOH = 0,03\,mol\\
nN{a_2}O = \dfrac{{0,09}}{2} = 0,045\,mol\\
a = 0,03 \times 23 + 0,045 \times 62 = 3,48g\\
c)\\
{C_M}NaOH = \dfrac{{0,12}}{{0,6}} = 0,2M
\end{array}\)