Đáp án:
\({m_{C{l_2}}}= 24,85{\text{ gam}}\)
\(V = 1,4{\text{ lít}}\)
\({m_{FeC{l_3}}}= 37,9167{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Mn{O_2} + 4HCl\xrightarrow{{}}MnC{l_2} + C{l_2} + 2{H_2}O\)
Ta có:
\({n_{Mn{O_2}}} = \frac{{30,45}}{{55 + 16.2}} = 0,35{\text{ mol = }}{{\text{n}}_{C{l_2}}}\)
\( \to {m_{C{l_2}}} = 0,35.71 = 24,85{\text{ gam}}\)
\({n_{HCl}} = 4{n_{C{l_2}}} = 1,4{\text{ mol}}\)
\( \to V = {V_{HCl}} = \frac{{1,4}}{1} = 1,4{\text{ lít}}\)
Phản ứng xảy ra:
\(2Fe + 3C{l_2}\xrightarrow{{{t^o}}}2FeC{l_3}\)
Ta có:
\({n_{Fe}} = \frac{{25,2}}{{56}} = 0,4{\text{ mol > }}\frac{2}{3}{n_{C{l_2}}}\)
Vậy \(Fe\) dư.
\( \to {n_{FeC{l_3}}} = \frac{2}{3}{n_{C{l_2}}} = \frac{{0,7}}{3}\)
\( \to {m_{FeC{l_3}}} = \frac{{0,7}}{3}.(56 + 35,3.3) = 37,9167{\text{ gam}}\)
