Đáp án:
\(\begin{array}{l}
a){m_{{{(C{H_3}{\rm{COO)}}}_2}Mg}} = 7,1g\\
b)C{\% _{{{(C{H_3}{\rm{COO)}}}_2}Mg}} = 3,5\% \\
c){m_{Mg{{(OH)}_2}}} = 2,9g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
MgO + 2C{H_3}{\rm{COO}}H \to {(C{H_3}{\rm{COO)}}_2}Mg + {H_2}O\\
{n_{MgO}} = 0,05mol\\
a)\\
{n_{{{(C{H_3}{\rm{COO)}}}_2}Mg}} = {n_{MgO}} = 0,05mol\\
\to {m_{{{(C{H_3}{\rm{COO)}}}_2}Mg}} = 7,1g
\end{array}\)
\(\begin{array}{l}
b)\\
{m_{{\rm{dd}}}} = {m_{MgO}} + {m_{{\rm{dd}}C{H_3}{\rm{COO}}H}} = 202g\\
\to C{\% _{{{(C{H_3}{\rm{COO)}}}_2}Mg}} = \dfrac{{7,1}}{{202}} \times 100\% = 3,5\%
\end{array}\)
\(\begin{array}{l}
c)\\
{(C{H_3}{\rm{COO)}}_2}Mg + 2KOH \to Mg{(OH)_2} + 2C{H_3}{\rm{COO}}K\\
{n_{KOH}} = 0,1mol\\
\to {n_{KOH}} > {n_{{{(C{H_3}{\rm{COO)}}}_2}Mg}} \to {n_{KOH}}dư\\
\to {n_{Mg{{(OH)}_2}}} = {n_{{{(C{H_3}{\rm{COO)}}}_2}Mg}} = 0,05mol\\
\to {m_{Mg{{(OH)}_2}}} = 2,9g
\end{array}\)
