Đáp án:
\({m_{HCl}}= 36,5{\text{ gam}}\)
\({m_{FeC{l_2}}} = 63,5{\text{ gam}}\)
\({V_{{H_2}}} = 11,2{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
\({n_{Fe}} = \frac{{28}}{{56}} = 0,5{\text{ mol = }}{{\text{n}}_{FeC{l_2}}} = {n_{{H_2}}}\)
\({n_{HCl}} = 2{n_{Fe}} = 0,5.2 = 1\;{\text{mol}}\)
\( \to {m_{HCl}} = 1.36,5 = 36,5{\text{ gam}}\)
\({m_{FeC{l_2}}} = 0,5.(56 + 35,5.2) = 63,5{\text{ gam}}\)
\({V_{{H_2}}} = 0,5.22,4 = 11,2{\text{ lít}}\)