Đáp án:

\(  {m_{dd{\text{ javen}}}} = 706,5{\text{ gam}}\)

Giải thích các bước giải:

 Các phản ứng xảy ra:

\(2NaCl + 2{H_2}O\xrightarrow{{đpdd/cmn}}2NaOH + C{l_2} + {H_2}\)

\(2NaOH + C{l_2}\xrightarrow{{}}NaCl + NaClO + {H_2}O\)

Ta có:

\({n_{C{l_2}}} = \frac{{33,6}}{{22,4}} = 1,5{\text{ mol}}\)

\({m_{NaOH}} = 600.20\%  = 120{\text{ gam}}\)

\( \to {n_{NaOH}} = \frac{{120}}{{40}} = 3{\text{ mol = 2}}{{\text{n}}_{C{l_2}}}\)

Vậy phản ứng vừa đủ

\( \to {m_{dd{\text{ javen}}}} = {m_{dd\;{\text{NaOH}}}} + {m_{dd\;{\text{C}}{{\text{l}}_2}}} = 600 + 1,5.71 = 706,5{\text{ gam}}\)

$n_{Cl_2}=\dfrac{33,6}{22,4}=1,5(mol)$

$\text{PTHH:}$

$2NaCl + 2H_2O\xrightarrow{\text{điện phân    ddcmn}} 2NaOH +Cl_2 +H_2$

$2NaOH + Cl_2\to NaClO+NaCl+H_2O$

$\to m_{Cl_2}=1,5.71=106,5g$

$\text{ Áp dụng ĐLBTKL, ta có:}$

$m_{\text{dd nước Javen}}=m_{m_{\text{dd HCl}}}+m_{\text{dd NaOH}}$

$\to m_{\text{ dd nước Javen}}=106,5+600=706,5g$