Đáp án:
\(\begin{array}{l}
b)\\
{m_{Mg}} = 3g\\
{m_{MgC{O_3}}} = 8,4g\\
c)\\
\% Mg = 26,3\% \\
\% MgC{O_3} = 73,7\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
MgC{O_3} + 2HCl \to MgC{l_2} + C{O_2} + {H_2}O\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
b)\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{2,8}}{{22,4}} = 0,125mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,125mol\\
{m_{Mg}} = n \times M = 0,125 \times 24 = 3g\\
{n_{CaC{O_3}}} = \dfrac{m}{M} = \dfrac{{10}}{{100}} = 0,1mol\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,1mol\\
\Rightarrow {n_{MgC{O_3}}} = {n_{C{O_2}}} = 0,1mol\\
{m_{MgC{O_3}}} = n \times M = 0,1 \times 84 = 8,4g\\
c)\\
m = {m_{Mg}} + {m_{MgC{O_3}}} = 3 + 8,4 = 11,4g\\
\% Mg = \dfrac{{{m_{Mg}}}}{m} \times 100\% = \dfrac{3}{{11,4}} \times 100\% = 26,3\% \\
\% MgC{O_3} = 100 - 26,3 = 73,7\%
\end{array}\)
