Đáp án:
a) % Fe phản ứng=83,33%
b)mFeCl3=10,833 gam ; mFeCl2=6,7733 gam
Giải thích các bước giải:
\(3Fe + 2{O_2}\xrightarrow{{}}F{e_3}{O_4}\)
\({n_{Fe}} = \frac{{6,72}}{{56}} = 0,12{\text{ mol}}\)
Cho hỗn hợp A tác dụng với HCl.
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
\(F{e_3}{O_4} + 8HCl\xrightarrow{{}}FeC{l_2} + 2FeC{l_3} + 4{H_2}O\)
\(\to {n_{Fe{\text{ dư}}}} = {n_{{H_{2\,}}}} = \frac{{0,448}}{{22,4}} = 0,02{\text{ mol}} \to {{\text{n}}_{Fe{\text{ phản ứng}}}} = 0,12 - 0,02 = 0,1{\text{ mol}}\)
\(\to {\% _{Fe{\text{ chuyển hoá}}}} = \frac{{0,1}}{{0,12}} = 83,33\% \)
\(\to {n_{F{e_3}{O_4}}} = \frac{1}{3}{n_{Fe{\text{ phan ung}}}} = \frac{{0,1}}{3} \to {n_{FeC{l_3}}} = 2{n_{F{e_3}{O_4}}} = \frac{{0,2}}{3}{\text{ mol}} \to {{\text{n}}_{FeC{l_2}}} = 0,12 - \frac{{0,2}}{3} = \frac{4}{{75}}{\text{ mol}}\)
\(\to {m_{FeC{l_3}}} = \frac{{0,2}}{3}.(56 + 35,5.3) = 10,833{\text{ gam; }}{{\text{m}}_{FeC{l_2}}} = \frac{4}{{75}}.(56 + 35,5.2) = 6,7733{\text{ gam}}\)