Đáp án:
\(\begin{array}{l}
a)\\
{m_{F{e_2}{O_3}}} = 10,4g\\
b)\\
{V_{\,{\rm{dd}}\,FeC{l_2}}} = 65\,ml\\
c)\\
{C_M}KCl = 2,36M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
FeC{l_2} + 2KOH \to Fe{(OH)_2} + 2KCl\\
4Fe{(OH)_2} + 2{H_2}O + {O_2} \to 4Fe{(OH)_3}\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O\\
{n_{KOH}} = \dfrac{{56 \times 26\% }}{{56}} = 0,26\,mol\\
{n_{Fe{{(OH)}_2}}} = \dfrac{{0,26}}{2} = 0,13\,mol\\
{n_{Fe{{(OH)}_3}}} = {n_{Fe{{(OH)}_2}}} = 0,13\,mol\\
{n_{F{e_2}{O_3}}} = \dfrac{{0,13}}{2} = 0,065\,mol\\
{m_{F{e_2}{O_3}}} = 0,065 \times 160 = 10,4g\\
b)\\
{n_{FeC{l_2}}} = {n_{Fe{{(OH)}_2}}} = 0,13\,mol\\
{V_{\,{\rm{dd}}\,FeC{l_2}}} = \dfrac{{0,13}}{2} = 0,065l = 65\,ml\\
c)\\
{V_{{\rm{dd}}KOH}} = \dfrac{{56}}{{1,24}} \approx 45\,ml\\
{V_{{\rm{dd}}spu}} = 0,065 + 0,045 = 0,11l\\
{n_{KCl}} = {n_{KOH}} = 0,26\,mol\\
{C_M}KCl = \dfrac{{0,26}}{{0,11}} = 2,36M
\end{array}\)
