Đáp án:
\({n_{HCl}} = 0,96{\text{ mol}}\)
\( \m={m_{KMn{O_4}}} = 18,96{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra :
\(2KMn{O_4} + 16HCl\xrightarrow{{}}2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\)
\(2Fe + 3C{l_2}\xrightarrow{{{t^o}}}2FeC{l_3}\)
Ta có:
\({n_{FeC{l_3}}} = \frac{{32,5}}{{56 + 35,5.3}} = 0,2{\text{ mol}}\)
\( \to {n_{C{l_2}}} = \frac{3}{2}{n_{FeC{l_3}}} = 0,3{\text{ mol}}\)
Ta có:
\({n_{HCl}} = \frac{{16}}{5}{n_{C{l_2}}} = 0,96{\text{ mol}}\)
\({n_{KMn{O_4}}} = \frac{2}{5}{n_{C{l_2}}} = 0,12{\text{ mol}}\)
\( \to m={m_{KMn{O_4}}} = 0,12.158 = 18,96{\text{ gam}}\)
