Đáp án:
2,9g
Giải thích các bước giải:
\(\begin{array}{l}
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
HCl + NaOH \to NaCl + {H_2}O\\
MgC{l_2} + 2NaOH \to Mg{(OH)_2} + 2NaCl\\
ZnC{l_2} + 2NaOH \to Zn{(OH)_2} + 2NaCl\\
Zn{(OH)_2} + 2NaOH \to N{a_2}Zn{O_2} + 2{H_2}O\\
{n_{Mg}} = \dfrac{{1,2}}{{24}} = 0,05\,mol\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,05\,mol\\
{n_{Mg{{(OH)}_2}}} = {n_{MgC{l_2}}} = 0,05\,mol\\
{m_{Mg{{(OH)}_2}}} = 0,05 \times 58 = 2,9g\\
\end{array}\)