Đáp án:
\( {C_{M{\text{ NaCl}}}} = {C_{M{\text{ NaClO}}}} = 1,6M\)
\({C_{M{\text{ NaOH}}}} = 0,8M\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Mn{O_2} + 4HCl\xrightarrow{{}}MnC{l_2} + C{l_2} + 2{H_2}O\)
Ta có:
\({n_{Mn{O_2}}} = \frac{{69,9}}{{55 + 16.2}} = 0,8{\text{ mol}}\)
\( \to {n_{C{l_2}}} = {n_{MgC{l_2}}} = 0,8{\text{ mol}}\)
\({n_{NaOH}} = 0,5.4 = 2{\text{ mol > 2}}{{\text{n}}_{C{l_2}}}\)
Do vậy \(NaOH\) dư.
\(2NaOH + C{l_2}\xrightarrow{{}}NaCl + NaClO + {H_2}O\)
\( \to {n_{NaCl}} = {n_{NaClO}} = {n_{C{l_2}}} = 0,8{\text{ mol}}\)
\({n_{NaOH}} = 2 - 0,8.2 = 0,4{\text{ mol}}\)
\( \to {C_{M{\text{ NaCl}}}} = {C_{M{\text{ NaClO}}}} = \frac{{0,8}}{{0,5}} = 1,6M\)
\({C_{M{\text{ NaOH}}}} = \frac{{0,4}}{{0,5}} = 0,8M\)
