\(\begin{array}{l}Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O\\C{l_2} + 2NaOH \to NaCl + NaClO + {H_2}O\end{array}\)
\({n_{C{l_2}}} = {n_{Mn{O_2}}} = \dfrac{{17,4}}{{87}} = 0,2\,mol\)
\(C{\% _{NaOH}} = \dfrac{{{m_{ct}}}}{{{m_{dd}}}} \times 100\% \Rightarrow {m_{NaOH}} = \dfrac{{C\% .{m_{{\text{dd}}}}}}{{100\% }} = \dfrac{{20.145,8}}{{100}} = 29,16\left( g \right)\)
\( \Rightarrow {n_{NaOH}} = \dfrac{{{m_{NaOH}}}}{{{M_{NaOH}}}} = \dfrac{{29,16}}{{40}} = 0,729\left( {mol} \right)\)
\(Cl_2\) + 2NaOH --> NaCl + NaClO + \(H_2O\)
Trước pứ:0,2 0,729
Trong pứ :0,2 ---> 0,4 -----------> 0,2 ------> 0,2 (mol)
Sau pứ : 0 -----> 0,329 -------> 0,2 ------> 0,2 (mol)
Số mol các chất sau phản ứng
\(n_{NaOH}\) = 0,329 mol => \(m_{NaOH}\) = 13,16 g
\(n_{NaCl}\) = 0,2 mol => \(m_{NaCl}\) = 11,7 g
\(n_{NaClO}\)= 0,2 mol => \(m_{NaClO}\) = 14,9 g
Vậy chứa các chất sau NaOH dư , NaCl , NaClO
\({m_{C{l_2}}} = 71.0,2 = 14,2\left( g \right) \Rightarrow {m_{{\text{dd}}saupu}} = {m_{C{l_2}}} + {m_{{\text{dd}}NaOH}} = 14,2 + 145,8 = 160\left( g \right)\)
C% NaOH = mct / mdd x 100% = 13,16 / 160 x 100% = 8,23 %
C% NaCl = mct / mdd x 100% = 11,7 / 160 x 100% = 7,31 %
C% NaClO = mct / mdd x 100% = 14,9 / 160 x 100% = 9,31 %
