Đáp án:
\( {m_{Fe}} = 5,6{\text{ gam}}\)
\({m_{F{e_2}{O_3}}} = 12{\text{ gam}}\)
\( {m_{hh}} = 17,6{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
\(F{e_2}{O_3} + 6HCl\xrightarrow{{}}2FeC{l_3} + 3{H_2}O\)
Ta có:
\({n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol = }}{{\text{n}}_{Fe}}\)
Ta có:
\({n_{FeS{O_4}}} = {n_{Fe}} = 0,1{\text{ mol}}\)
\( \to {m_{FeS{O_4}}} = 0,1.152 = 15,2{\text{ gam}}\)
\( \to {m_{F{e_2}{{(S{O_4})}_3}}} = 45,2 - 15,2 = 30{\text{ gam}}\)
\( \to {n_{F{e_2}{{(S{O_4})}_3}}} = \frac{{30}}{{56.2 + 96.3}} = 0,075{\text{ mol}}\)
\( \to {n_{F{e_2}{O_3}}} = {n_{F{e_2}{{(S{O_4})}_3}}} = 0,075{\text{ mol}}\)
\( \to {m_{Fe}} = 0,1.56 = 5,6{\text{ gam}}\)
\({m_{F{e_2}{O_3}}} = 0,075.160 = 12{\text{ gam}}\)
\( \to {m_{hh}} = 12 + 5,6 = 17,6{\text{ gam}}\)
