Đáp án:
Bạn tham khảo lời giải ở dưới nhé!!
Giải thích các bước giải:
7,
a,
\(\begin{array}{l}
{n_{{K_2}S{O_4}}} = 0,075mol\\
{n_{N{a_2}S{O_4}}} = 0,3mol\\
\to {n_{{K^ + }}} = 2{n_{{K_2}S{O_4}}} = 0,15mol\\
\to {n_{N{a^ + }}} = 2{n_{N{a_2}S{O_4}}} = 0,6mol\\
\to {n_{S{O_4}^{2 - }}} = {n_{{K_2}S{O_4}}} + {n_{N{a_2}S{O_4}}} = 0,375mol\\
\to C{M_{{K^ + }}} = \dfrac{{0,15}}{{0,3}} = 0,5M\\
\to C{M_{N{a^ + }}} = \dfrac{{0,6}}{{0,3}} = 2M\\
\to C{M_{S{O_4}^{2 - }}} = \dfrac{{0,375}}{{0,3}} = 1,25M
\end{array}\)
b,
\(\begin{array}{l}
FeC{l_3} + 3KOH \to Fe{(OH)_3} + 3KCl\\
{n_{FeC{l_3}}} = 0,01mol\\
{n_{KOH}} = 0,8mol\\
{n_{FeC{l_3}}} < \dfrac{{{n_{KOH}}}}{3}
\end{array}\)
Suy ra KOH dư
\(\begin{array}{l}
\to {n_{KOH}}dư= 0,8 - 0,01 \times 3 = 0,77mol\\
\to {n_{KCl}} = 3{n_{FeC{l_3}}} = 0,03mol\\
\to {n_{{K^ + }}} = {n_{KOH}}dư+ {n_{KCl}} = 0,8mol\\
\to {n_{O{H^ - }}} = {n_{KOH}}dư= 0,77mol\\
\to {n_{C{l^ - }}} = {n_{KCl}} = 0,03mol\\
\to C{M_{{K^ + }}} = \dfrac{{0,8}}{{0,5}} = 1,6M\\
\to C{M_{O{H^ - }}} = \dfrac{{0,77}}{{0,5}} = 1,54M\\
\to C{M_{C{l^ - }}} = \dfrac{{0,03}}{{0,5}} = 0,06M
\end{array}\)
c,
\(\begin{array}{l}
{n_{HCl}} = 0,045mol\\
{n_{HN{O_3}}} = 0,04mol\\
\to {n_{{H^ + }}} = {n_{HCl}} + {n_{HN{O_3}}} = 0,085mol\\
\to {n_{C{l^ - }}} = {n_{HCl}} = 0,045mol\\
\to {n_{N{O_3}^ - }} = {n_{HN{O_3}}} = 0,04mol\\
\to C{M_{{H^ + }}} = \dfrac{{0,085}}{{0,5}} = 0,17M\\
\to C{M_{C{l^ - }}} = \dfrac{{0,045}}{{0,5}} = 0,09M\\
\to C{M_{N{O_3}^ - }} = \dfrac{{0,04}}{{0,5}} = 0,08M
\end{array}\)
d,
\(\begin{array}{l}
{n_{NaOH}} = 0,1mol\\
{n_{Ba{{(OH)}_2}}} = 0,09mol\\
\to {n_{N{a^ + }}} = {n_{NaOH}} = 0,1mol\\
\to {n_{B{a^{2 + }}}} = {n_{Ba{{(OH)}_2}}} = 0,09mol\\
\to {n_{O{H^ - }}} = {n_{NaOH}} + 2{n_{Ba{{(OH)}_2}}} = 0,28mol\\
\to C{M_{N{a^ + }}} = \dfrac{{0,1}}{{0,5}} = 0,2M\\
\to C{M_{B{a^{2 + }}}} = \dfrac{{0,09}}{{0,5}} = 0,18M\\
\to C{M_{O{H^ - }}} = \dfrac{{0,28}}{{0,5}} = 0,56M
\end{array}\)
e,
\(\begin{array}{l}
Ba{(OH)_2} + 2HN{O_3} \to Ba{(N{O_3})_2} + 2{H_2}O\\
{n_{Ba{{(OH)}_2}}} = 0,03mol\\
{n_{HN{O_3}}} = 0,04mol\\
{n_{Ba{{(OH)}_2}}} > \dfrac{{{n_{HN{O_3}}}}}{2}
\end{array}\)
Suy ra \(Ba{(OH)_2}\) dư
\(\begin{array}{l}
\to {n_{Ba{{(OH)}_2}}}dư= 0,03 - \dfrac{1}{2} \times 0,04 = 0,01mol\\
\to {n_{Ba{{(N{O_3})}_2}}} = \dfrac{1}{2}{n_{HN{O_3}}} = 0,02mol\\
\to {n_{B{a^{2 + }}}} = {n_{Ba{{(OH)}_2}}}dư+ {n_{Ba{{(N{O_3})}_2}}} = 0,03mol\\
\to {n_{O{H^ - }}} = 2{n_{Ba{{(OH)}_2}}}dư= 0,02mol\\
\to {n_{N{O_3}^ - }} = 2{n_{Ba{{(N{O_3})}_2}}} = 0,04mol\\
\to C{M_{B{a^{2 + }}}} = \dfrac{{0,03}}{{0,4}} = 0,075M\\
\to C{M_{O{H^ - }}} = \dfrac{{0,02}}{{0,4}} = 0,05M\\
\to C{M_{N{O_3}^ - }} = \dfrac{{0,04}}{{0,4}} = 0,1M
\end{array}\)
