Đáp án:
\( \to {C_{M{\text{ NaCl}}}} = {C_{M{\text{ NaClO}}}} = 1,65M;{C_{M{\text{ NaOH}}}} = 0,7M\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2KMn{O_4} + 16HCl\xrightarrow{{}}2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\)
\(2NaOH + C{l_2}\xrightarrow{{}}NaCl + NaClO + {H_2}O\)
Ta có:
\({n_{KMn{O_4}}} = \frac{{52,2}}{{39 + 55 + 16.4}} = 0,33{\text{ mol}}\)
\( \to {n_{C{l_2}}} = \frac{5}{2}{n_{KMn{O_4}}} = 0,825{\text{ mol}}\)
\({n_{NaOH}} = 0,5.4 = 2{\text{ mol}}\)
\( \to \frac{{{n_{NaOH}}}}{{{n_{C{l_2}}}}} = \frac{2}{{0,825}} > 2\) nên \(NaOH\) dư
\( \to {n_{NaCl}} = {n_{NaClO}} = {n_{C{l_2}}} = 0,825{\text{ mol}}\)
\({n_{NaOH{\text{ dư}}}} = 2 - 0,825.2 = 0,35{\text{ mol}}\)
\( \to {C_{M{\text{ NaCl}}}} = {C_{M{\text{ NaClO}}}} = \frac{{0,825}}{{0,5}} = 1,65M;{C_{M{\text{ NaOH}}}} = \frac{{0,35}}{{0,5}} = 0,7M\)
