Cho hỗn hợp Y qua 40 gam dung dịch KOH 14% nung nóng được dung dịch Z. Khối lượng muối clorua trong Z là

Số mol KOH ban đầu: \(C\%  = \dfrac{{{m_{KOH}}}}{{{m_{{\rm{dd}}KOH}}}}.100\%  \Leftrightarrow 14\%  = \dfrac{{56.{n_{KOH}}}}{{40}}.100\%  \Leftrightarrow {n_{KOH}} = 0,1mol\)

\(\begin{array}{*{20}{c}}{HCl}\\{0,06}\end{array}\begin{array}{*{20}{c}} + \\{}\end{array}\begin{array}{*{20}{c}}{KOH}\\{0,06}\end{array}\begin{array}{*{20}{c}} \to \\{}\end{array}\begin{array}{*{20}{c}}{KCl}\\{0,06}\end{array}\begin{array}{*{20}{c}} + \\{}\end{array}\begin{array}{*{20}{c}}{{H_2}O}\\{}\end{array}\)

\(\begin{array}{*{20}{c}}{3C{l_2}}\\{0,0075}\end{array}\begin{array}{*{20}{c}} + \\{}\end{array}\begin{array}{*{20}{c}}{6KOH}\\{0,015}\end{array}\begin{array}{*{20}{c}} \to \\{}\end{array}\begin{array}{*{20}{c}}{5KCl}\\{0,0125}\end{array}\begin{array}{*{20}{c}} + \\{}\end{array}\begin{array}{*{20}{c}}{KCl{O_3}}\\{}\end{array}\begin{array}{*{20}{c}} + \\{}\end{array}\begin{array}{*{20}{c}}{3{H_2}O}\\{}\end{array}\)

Áp dụng BTKL:

Khối lượng dung dịch Z = \({m_{C{l_2}}} + {m_{HCl}} + {m_{{\rm{dd}}KOH}} = 0,0075.71 + 0,06.36,5 + 40 = 42,7225g\)

\({C_{\% (KCl)}} = \dfrac{{{m_{KCl}}}}{{{m_{{\rm{dd}}Z}}}}.100\%  = \dfrac{{(0,06 + 0,0125).74,5}}{{42,7225}}.100\%  \simeq 12,64\% \)