Đáp án:
\(\begin{array}{l}
b)\\
\% Mg = 48\% \\
\% MgO = 52\% \\
c)\\
C{\% _{HCl}} = 12,045\% \\
d)\\
C{\% _{MgC{l_2}}} = 14,96\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
b)\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,2mol\\
{m_{Mg}} = n \times M = 0,2 \times 24 = 4,8g\\
{m_{MgO}} = 10 - 4,8 = 5,2g\\
\% Mg = \dfrac{{4,8}}{{10}} \times 100\% = 48\% \\
\% MgO = 100 - 48 = 52\% \\
c)\\
{n_{MgO}} = \dfrac{m}{M} = \dfrac{{5,2}}{{40}} = 0,13mol\\
{n_{HCl}} = 2{n_{Mg}} + 2{n_{MgO}} = 0,66mol\\
{m_{HCl}} = n \times M = 0,66 \times 36,5 = 24,09g\\
C{\% _{HCl}} = \dfrac{{24,09}}{{200}} \times 100\% = 12,045\% \\
d)\\
{n_{MgC{l_2}}} = {n_{Mg}} + {n_{MgO}} = 0,33mol\\
{m_{MgC{l_2}}} = n \times M = 0,33 \times 95 = 31,35g\\
{m_{{\rm{dd}}spu}} = 10 + 200 - 0,2 \times 2 = 209,6g\\
C{\% _{MgC{l_2}}} = \dfrac{{31,35}}{{209,6}} \times 100\% = 14,96\%
\end{array}\)
