Đáp án:
\(\begin{array}{l}
2)\\
a)\\
{C_{M{\rm{dd}}\,A}} = 3M\\
b)\\
a = 16g\\
{C_M}F{e_2}{(S{O_4})_3} = 0,4M\\
{C_M}N{a_2}S{O_4} = 0,6M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2)\\
a)\\
N{a_2}O + {H_2}O \to 2NaOH\\
{n_{N{a_2}O}} = \dfrac{{18,6}}{{62}} = 0,3\,mol\\
{n_{NaOH}} = 2{n_{N{a_2}O}} = 0,6\,mol\\
{C_{M{\rm{dd}}\,A}} = \dfrac{{0,6}}{{0,2}} = 3M\\
b)\\
F{e_2}{(S{O_4})_3} + 6NaOH \to 2Fe{(OH)_3} + 3N{a_2}S{O_4}\\
{n_{F{e_2}{{(S{O_4})}_3}}} = 0,3 \times 1 = 0,3\,mol\\
\dfrac{{0,3}}{1} > \dfrac{{0,6}}{6} \Rightarrow\text{ $Fe_2(SO_4)_3$ dư} \\
{n_{Fe{{(OH)}_3}}} = \dfrac{{0,6 \times 2}}{6} = 0,2\,mol\\
2Fe{(OH)_3} \xrightarrow{t^0} F{e_2}{O_3} + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = \dfrac{{0,2}}{2} = 0,1\,mol\\
a = {m_{F{e_2}{O_3}}} = 0,1 \times 160 = 16g\\
{n_{F{e_2}{{(S{O_4})}_3}}} \text{ dư}= 0,3 - \dfrac{{0,6}}{6} = 0,2\,mol\\
{V_{{\rm{dd}}}} = 0,3 + 0,2 = 0,5l\\
{n_{N{a_2}S{O_4}}} = \dfrac{{0,6}}{2} = 0,3\,mol\\
{C_M}F{e_2}{(S{O_4})_3} = \dfrac{{0,2}}{{0,5}} = 0,4M\\
{C_M}N{a_2}S{O_4} = \dfrac{{0,3}}{{0,5}} = 0,6M
\end{array}\)
