Đáp án:
\(\begin{array}{l}
a)\\
{m_{Mg}} = 2,4g\\
{m_{MgO}} = 8g\\
c)\\
\% {m_{Mg}} = 23,08\% \\
\% {m_{MgO}} = 76,92\% \\
d)\\
{C_\% }MgS{O_4} = 11,61\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
MgO + {H_2}S{O_4} \to MgS{O_4} + {H_2}O\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,1\,mol\\
{m_{Mg}} = 0,1 \times 24 = 2,4g\\
{m_{MgO}} = 10,4 - 2,4 = 8g\\
c)\\
\% {m_{Mg}} = \dfrac{{2,4}}{{10,4}} \times 100\% = 23,08\% \\
\% {m_{MgO}} = 100 - 23,08 = 76,92\% \\
d)\\
{m_{{\rm{dd}}spu}} = 10,4 + 300 - 0,1 \times 2 = 310,2g\\
{n_{MgO}} = \dfrac{8}{{40}} = 0,2\,mol\\
{n_{MgS{O_4}}} = 0,2 + 0,1 = 0,3\,mol\\
{C_\% }MgS{O_4} = \dfrac{{0,3 \times 120}}{{310,2}} \times 100\% = 11,61\%
\end{array}\)
