Đáp án:
b)\({V_{C{l_2}}} = 5,6{\text{ lít}}\)
c) \({C_{M{\text{ NaCl}}}} = {C_{M{\text{ NaClO}}}} = 0,4167M;\;{{\text{C}}_{M{\text{ NaOH dư}}}} = 1,167M\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2KMn{O_4} + 16HCl\xrightarrow{{}}2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\)
Ta có:
\({n_{KMn{O_4}}} = \frac{{15,8}}{{39 + 55 + 16.4}} = 0,1{\text{ mol}} \to {{\text{n}}_{C{l_2}}} = \frac{5}{2}{n_{KMn{O_4}}} = 0,25{\text{ mol}} \to {{\text{V}}_{C{l_2}}} = 0,25.22,4 = 5,6{\text{ lít}}\)
\(C{l_2} + 2NaOH\xrightarrow{{}}NaCl + NaClO + {H_2}O\)
Ta có: \({n_{NaOH}} = 0,6.2 = 1,2{\text{ mol > 2}}{{\text{n}}_{C{L_2}}}\) do vậy NaOH dư.
\(\to {n_{NaCl}} = {n_{NaClO}} = 0,25{\text{ mol; }}{{\text{n}}_{NaOH{\text{ dư}}}} = 1,2 - 0,25.2 = 0,7{\text{ mol}}\)
\(\to {C_{M{\text{ NaCl}}}} = {C_{M{\text{ NaClO}}}} = \frac{{0,25}}{{0,6}} = 0,4167M;\;{{\text{C}}_{M{\text{ NaOH du}}}} = \frac{{0,7}}{{0,6}} = 1,167M\)