Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Mg}} = 37,5\% \\
\% {m_{MgO}} = 62,5\% \\
b)\\
{m_{{\rm{dd}}HCl}} = 200g\\
c)\\
{C_\% }MgC{l_2} = 13,62\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,15\,mol\\
\% {m_{Mg}} = \dfrac{{0,15 \times 24}}{{9,6}} \times 100\% = 37,5\% \\
\% {m_{MgO}} = 100 - 37,5 = 62,5\% \\
b)\\
{n_{MgO}} = \dfrac{{9,6 - 0,15 \times 24}}{{40}} = 0,15\,mol\\
{n_{HCl}} = 0,15 \times 2 + 0,15 \times 2 = 0,6\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,6 \times 36,5}}{{10,95\% }} = 200g\\
c)\\
{m_{{\rm{dd}}spu}} = 9,6 + 200 - 0,15 \times 2 = 209,3g\\
{n_{MgC{l_2}}} = 0,15 + 0,15 = 0,3\,mol\\
{C_\% }MgC{l_2} = \dfrac{{0,3 \times 95}}{{209,3}} \times 100\% = 13,62\%
\end{array}\)