Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Mg}} = 28,57\% \\
\% {m_{MgO}} = 71,43\% \\
b)\\
{m_{{\rm{dd}}HCl}} = 500g\\
{C_\% }MgC{l_2} = 4,67\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,1\,mol\\
\% {m_{Mg}} = \dfrac{{0,1 \times 24}}{{8,4}} \times 100\% = 28,57\% \\
\% {m_{MgO}} = 100 - 28,57 = 71,43\% \\
b)\\
{n_{MgO}} = \dfrac{6}{{40}} = 0,15\,mol\\
{n_{HCl}} = 0,1 \times 2 + 0,15 \times 2 = 0,5\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,5 \times 36,5}}{{3,65\% }} = 500g\\
{m_{{\rm{dd}}Y}} = 8,4 + 500 - 0,1 \times 2 = 508,2g\\
{n_{MgC{l_2}}} = 0,1 + 0,15 = 0,25\,mol\\
{C_\% }MgC{l_2} = \dfrac{{0,25 \times 95}}{{508,2}} \times 100\% = 4,67\%
\end{array}\)
