Đáp án:
CM KCl= CM KClO=0,9M; CM KOH dư=0,2M
Giải thích các bước giải:
\(Mn{O_2} + 4HCl\xrightarrow{{}}MnC{l_2} + C{l_2} + 2{H_2}O\)
Ta có: \({n_{Mn{O_2}}} = \frac{{15,66}}{{55 + 16.2}} = 0,18{\text{ mol}}\)
Theo phản ứng: \({n_{C{l_2}}} = {n_{Mn{O_2}}} = 0,18{\text{ mol}}\)
Ta có: \({n_{KOH}} = 0,2.2 = 0,4{\text{ mol}}\)
\(2KOH + C{l_2}\xrightarrow{{}}KCl + KClO + {H_2}O\)
\(\to {n_{KCl}} = {n_{KClO}} = {n_{C{l_2}}} = 0,18{\text{ mol}} \to {{\text{n}}_{KOH{\text{ dư}}}} = 0,4 - 0,18.2 = 0,04{\text{ mol}}\)
\(\to {C_{M{\text{ }}KCl}} = {C_{M{\text{ KClO}}}} = \frac{{0,18}}{{0,2}} = 0,9M;{\text{ }}{{\text{C}}_{M{\text{ KOH dư}}}} = \frac{{0,04}}{{0,2}} = 0,2M\)