Đáp án:
B
Giải thích các bước giải:
\(2KMn{O_4} + 16HCl\xrightarrow{{}}2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\)
\(2Fe + 3C{l_2}\xrightarrow{{}}2FeC{l_3}\)
Ta có: \({n_{FeC{l_3}}} = \frac{{32,5}}{{56 + 35,5.3}} = 0,2{\text{ mol}} \to {{\text{n}}_{C{l_2}}} = \frac{3}{2}{n_{FeC{l_3}}} = 0,3{\text{ mol}}\)
Theo phản ứng:
\({n_{KMn{O_4}}} = \frac{2}{5}{n_{C{l_2}}} = 0,12{\text{ mol}} \to {{\text{m}}_{KMn{O_4}}} = 0,12.(39 + 55 + 16.4) = 18,96{\text{ gam}}\)
\({n_{HCl}} = \frac{{16}}{5}{n_{C{l_2}}} = 0,96{\text{ mol}} \to {{\text{V}}_{HCl}} = \frac{{0,96}}{1} = 0,96{\text{ lít = 960 ml}}\)