Đáp án:
a. $m = 103,96 gam$
b. $V=3,68 lit$
c. \({C_{M{\text{ FeC}}{{\text{l}}_2}}} = 0,0625M;\\{C_{M{\text{ FeC}}{{\text{l}}_3}}} = 0,125M\)
Giải thích các bước giải:
Phương trình hóa học:
\(Fe_3O_4+8HCl \rightarrow 2FeCl_3+FeCl_2+4H_2O\)
Ta có:
\({n_{F{e_3}{O_4}}} = \dfrac{{53,36}}{{56.3 + 16.4}} = 0,23{\text{ mol}}\)
\( \to {n_{FeC{l_3}}} = 2{n_{F{e_3}{O_4}}} = 0,46{\text{ mol;}}{{\text{n}}_{FeC{l_2}}} = {n_{F{e_3}{O_4}}} = 0,23 mol\)
\( \to {m_{muối}} = 0,46.(56 + 35,5.3) + 0,23.(56 + 35,5.2) \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,= 103,96{\text{ gam}}\)
\({n_{HCl}} = 8{n_{F{e_3}{O_4}}} = 0,23.8 = 1,84{\text{ gam}}\)
\( \to {V_{dd{\text{HCl}}}} = \dfrac{{1,84}}{{0,5}} = 3,68{\text{ lít = }}{{\text{V}}_{dd{\text{ muối}}}}\)
\({C_{M{\text{ FeC}}{{\text{l}}_2}}} = \dfrac{{0,23}}{{3,68}} = 0,0625M;{C_{M{\text{ FeC}}{{\text{l}}_3}}} = \dfrac{{0,46}}{{3,68}} = 0,125M\)
