Đáp án:
\({V_{C{l_2}}} = 11,2{\text{ lít}}\)
\({m_{muối}} = 71,7{\text{ gam}}\)
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:
\({n_{KMn{O_4}}} = \frac{{63,2}}{{39 + 55 + 16.4}} = 0,4{\text{ mol;}}{{\text{n}}_{HCl}} = 1,6.1 = 1,6{\text{ mol}}\)
Vì \({n_{HCl}} < 8{n_{KMn{O_4}}}\) nên \(KMnO_4\) dư.
\( \to {n_{C{l_2}}} = \frac{5}{{16}}{n_{HCl}} = 0,5{\text{ mol}}\)
\( \to {V_{C{l_2}}} = 0,5.22,4 = 11,2{\text{ lít}}\)
\({n_{KCl}} = {n_{MnC{l_2}}} = \frac{1}{8}{n_{HCl}} = 0,2{\text{ mol}}\)
\({n_{KMn{O_4}{\text{ dư}}}} = 0,4 - \frac{{1,6}}{8} = 0,2{\text{ mol}}\)
\( \to {m_{muối}} = {m_{KCl}} + {m_{MnC{l_2}}} + {m_{KMn{O_4}}}\)
\( = 0,2.74,5 + 0,2.126 + 0,2.158 = 71,7{\text{ gam}}\)
