Đáp án:
\(\begin{array}{l}
a){m_{Mg{{(OH)}_2}}} = 11,6g\\
b)\\
C{\% _{MgS{O_4}(dư)}} = 2,46\% \\
C{\% _{N{a_2}S{O_4}}} = 5,81\% \\
c){m_{MgO}} = 8g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
MgS{O_4} + 2NaOH \to Mg{(OH)_2} + N{a_2}S{O_4}\\
{m_{MgS{O_4}}} = \dfrac{{300 \times 12\% }}{{100\% }} = 36g\\
\to {n_{MgS{O_4}}} = 0,3mol\\
{m_{NaOH}} = \dfrac{{200 \times 8\% }}{{100\% }} = 16g\\
\to {n_{NaOH}} = 0,4mol\\
\to {n_{MgS{O_4}}} > \dfrac{{{n_{NaOH}}}}{2} \to {n_{MgS{O_4}}}dư
\end{array}\)
\(\begin{array}{l}
a)\\
{n_{Mg{{(OH)}_2}}} = \dfrac{1}{2}{n_{NaOH}} = 0,2mol\\
\to {m_{Mg{{(OH)}_2}}} = 11,6g
\end{array}\)
\(\begin{array}{l}
b)\\
{n_{N{a_2}S{O_4}}} = {n_{MgS{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,2mol\\
\to {n_{MgS{O_4}(dư)}} = 0,3 - 0,2 = 0,1mol\\
\to {m_{MgS{O_4}(dư)}} = 12g\\
\to {m_{N{a_2}S{O_4}}} = 28,4g\\
{m_{{\rm{dd}}}} = {m_{{\rm{dd}}MgS{O_4}}} + {m_{{\rm{dd}}NaOH}} - {m_{Mg{{(OH)}_2}}} = 488,4g\\
\to C{\% _{MgS{O_4}(dư)}} = \dfrac{{12}}{{488,4}} \times 100\% = 2,46\% \\
\to C{\% _{N{a_2}S{O_4}}} = \dfrac{{28,4}}{{488,4}} \times 100\% = 5,81\%
\end{array}\)
\(\begin{array}{l}
c)\\
Mg{(OH)_2} \to MgO + {H_2}O\\
{n_{MgO}} = {n_{Mg{{(OH)}_2}}} = 0,2mol\\
\to {m_{MgO}} = 8g
\end{array}\)
