\(\begin{array}{l}
7)\\
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
b)\\
nMg = \dfrac{m}{M} = \dfrac{{48}}{{24}} = 2\,mol\\
nHCl = 2nMg = 4\,mol\\
mHCl = n \times M = 4 \times 36,5 = 146g\\
8)\\
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
b)\\
nZn = \dfrac{m}{M} = \dfrac{{32,5}}{{65}} = 0,5\,mol\\
nHCl = 2nZn = 0,5 \times 2 = 1\,mol\\
mHCl = n \times M = 1 \times 36,5 = 36,5g\\
c)\\
{C_M}HCl = \dfrac{n}{V} = \dfrac{1}{{0,3}} = 3,33M
\end{array}\)
