Đáp án:

 16g

Giải thích các bước giải:

\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,2\,mol\\
\% {m_{Mg}} = \dfrac{{0,2 \times 24}}{{12}} \times 100\%  = 40\% \\
\% {m_{FeO}} = 100 - 40 = 60\% \\
{n_{HCl}} = 0,4 \times 2 = 0,8\,mol\\
{n_{HCl}} \text{ dư}= 0,8 - 0,2 \times 2 = 0,4\,mol\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{NaOH}} = {n_{HCl}} \text{ dư} = 0,4\,mol\\
{V_{NaOH}} = \dfrac{{0,4}}{{0,5}} = 0,8l\\
2NaOH + MgC{l_2} \to Mg{(OH)_2} + 2NaCl\\
FeC{l_2} + 2NaOH \to Fe{(OH)_2} + 2NaCl\\
4Fe{(OH)_2} + {O_2} + 2{H_2}O \to 4Fe{(OH)_3}\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O\\
Mg{(OH)_2} \to MgO + {H_2}O\\
{n_{FeO}} = \dfrac{{12 - 4,8}}{{72}} = 0,1\,mol\\
{n_{F{e_2}{O_3}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{n_{MgO}} = {n_{Mg}} = 0,2\,mol\\
m = 0,05 \times 160 + 0,2 \times 40 = 16g
\end{array}\)