Đáp án:

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Giải thích các bước giải:

\(\begin{array}{l}
1,\\
{K_2}S{O_3} + 2HCl \to 2KCl + S{O_2} + {H_2}O\\
{n_{{K_2}S{O_3}}} = \dfrac{{150 \times 15,8}}{{100 \times 158}} = 0,15mol\\
 \to {n_{HCl}} = 2{n_{{K_2}S{O_3}}} = 0,3mol\\
 \to {m_{HCl}} = 10,95g\\
 \to {m_{HCl}}{\rm{dd}} = \dfrac{{10,95 \times 100}}{{36,5}} = 30g\\
{n_{S{O_2}}} = {n_{{K_2}S{O_3}}} = 0,15mol\\
{n_{KCl}} = 2{n_{{K_2}S{O_3}}} = 0,3mol\\
{m_{{\rm{dd}}}} = 150 + 30 - 0,15 \times 64 = 170,4g\\
 \to C{\% _{KCl}} = \dfrac{{0,3 \times 74,5}}{{170,4}} \times 100\%  = 13,12\% 
\end{array}\)

\(\begin{array}{l}
2,\\
F{\rm{eC}}{{\rm{l}}_3} + 3K{\rm{O}}H \to F{\rm{e}}{(OH)_3} + 3KCl\\
{n_{F{\rm{e}}C{l_3}}} = \dfrac{{50 \times 16,25}}{{100 \times 162,5}} = 0,05mol\\
 \to {n_{F{\rm{e}}{{(OH)}_3}}} = {n_{F{\rm{e}}C{l_3}}} = 0,05mol\\
 \to {m_{F{\rm{e}}{{(OH)}_3}}} = 5,35g\\
{n_{K{\rm{O}}H}} = 3{n_{F{\rm{e}}C{l_3}}} = 0,15mol\\
 \to {m_{K{\rm{O}}H}} = 8,4g\\
 \to C{\% _{K{\rm{O}}H}} = \dfrac{{8,4}}{{50}} \times 100\%  = 16,8\% \\
{n_{KCl}} = {n_{K{\rm{O}}H}} = 0,15mol\\
{m_{{\rm{dd}}}} = 50 + 50 - 5,35 = 94,65g\\
 \to C{\% _{KCl}} = \dfrac{{0,15 \times 74,5}}{{94,65}} \times 100\%  = 11,81\% 
\end{array}\)

\(\begin{array}{l}
3,\\
Cu{(OH)_2} \to CuO + {H_2}O\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
{n_{Cu{{(OH)}_2}}} = 0,12mol\\
 \to {n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,12mol\\
 \to {m_A} = {m_{CuO}} = 9,6g\\
{n_{{H_2}S{O_4}}} = {n_{CuO}} = 0,12mol\\
 \to C{M_{{H_2}S{O_4}}} = \dfrac{{0,12}}{{0,1}} = 1,2M\\
{n_{CuS{O_4}}} = {n_{CuO}} = 0,12mol\\
 \to C{M_{CuS{O_4}}} = \dfrac{{0,12}}{{0,1}} = 1,2M
\end{array}\)

\(\begin{array}{l}
4,\\
AlC{l_3} + 3AgN{O_3} \to Al{(N{O_3})_3} + 3AgCl\\
{n_{AlC{l_3}}} = 0,05mol\\
{n_{AgN{O_3}}} = 0,06mol\\
 \to {n_{AlC{l_3}}} > \dfrac{{{n_{AgN{O_3}}}}}{3}
\end{array}\)

Suy ra \(AlC{l_3}\) dư 

\(\begin{array}{l}
 \to {n_{AgCl}} = {n_{AgN{O_3}}} = 0,06mol\\
 \to {m_{AgCl}} = 8,61g\\
{n_{Al{{(N{O_3})}_3}}} = \dfrac{1}{3}{n_{AgN{O_3}}} = 0,02mol\\
{n_{AlC{l_3}}}dư= 0,05 - \dfrac{1}{3} \times 0,06 = 0,03mol\\
 \to C{M_{AlC{l_3}}}dư= \dfrac{{0,03}}{{0,05 + 0,05}} = 0,3M\\
 \to C{M_{Al{{(N{O_3})}_3}}} = \dfrac{{0,02}}{{0,05 + 0,05}} = 0,2M
\end{array}\)

5,

\(\begin{array}{l}
CaC{l_2} + {H_2}S{O_4} \to CaS{O_4} + 2HCl\\
{n_{CaC{l_2}}} = \dfrac{{30 \times 4,44}}{{100 \times 111}} = 0,012mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{20 \times 7,95}}{{100 \times 98}} = 0,016mol\\
 \to {n_{CaC{l_2}}} < {n_{{H_2}S{O_4}}}
\end{array}\)

Suy ra \({H_2}S{O_4}\) dư

\(\begin{array}{l}
 \to {n_{{H_2}S{O_4}}}dư= 0,016 - 0,012 = 0,004mol\\
 \to {n_{CaS{O_4}}} = {n_{CaC{l_2}}} = 0,012mol\\
 \to {m_{CaS{O_4}}} = 1,632g\\
{n_{HCl}} = 2{n_{CaC{l_2}}} = 0,024mol\\
{m_{dd}} = 30 + 20 - 1,632 = 48,368g\\
 \to C{\% _{{H_2}S{O_4}}}dư= \dfrac{{0,004 \times 98}}{{48,368}} \times 100\%  = 0,81\% \\
 \to C{\% _{HCl}} = \dfrac{{0,024 \times 36,5}}{{48,368}} \times 100\%  = 1,81\% 
\end{array}\)