Đáp án:
a)
\(\begin{array}{l}
\% Mg = 13,04\% \\
\% MgO = 86,96\% \\
b)\\
{m_{NaOH}} = 4g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
\Rightarrow {n_{Mg}} = 0,05mol\\
{m_{Mg}} = n \times M = 0,05 \times 24 = 1,2g\\
{m_{MgO}} = m - {m_{Mg}} = 9,2 - 1,2 = 8g\\
\% Mg = \dfrac{{{m_{Mg}}}}{m} \times 100\% = \dfrac{{1,2}}{{9,2}} \times 100\% = 13,04\% \\
\% MgO = 100 - 13,04 = 86,96\% \\
b)\\
{n_{MgO}} = \dfrac{m}{M} = \dfrac{8}{{40}} = 0,2mol\\
{n_{HCl}} = V \times {C_M} = 0,2 \times 3 = 0,6mol\\
{n_{HC{l_d}}} = {n_{HCl}} - 2{n_{Mg}} - 2{n_{MgO}}\\
\,\,\,\,\,\,\,\,\,\, = 0,6 - 2 \times 0,05 - 2 \times 0,2 = 0,1mol\\
HCl + NaOH \to NaCl + {H_2}O\\
\Rightarrow {n_{NaOH}} = {n_{NaC{l_d}}} = 0,1mol\\
{m_{NaOH}} = n \times M = 0,1 \times 40 = 4g
\end{array}\)
