Đáp án:
\(a = 32{\text{ gam}}\)
\(b = 80{\text{ gam}}\)
\(C{\% _{F{e_2}{{(S{O_4})}_3}}} = 24,1\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(F{e_2}{O_3} + 3{H_2}S{O_4}\xrightarrow{{}}F{e_2}{(S{O_4})_3} + 3{H_2}O\)
Ta có:
\({m_{{H_2}S{O_4}}} = 300.19,6\% = 58,8{\text{ gam}}\)
\( \to {n_{{H_2}S{O_4}}} = \frac{{58,8}}{{98}} = 0,6{\text{ mol}}\)
\( \to {n_{F{e_2}{O_3}}} = {n_{F{e_2}{{(S{O_4})}_3}}} = \frac{1}{3}{n_{{H_2}S{O_4}}} = \frac{1}{3}.0,6 = 0,2{\text{ mol}}\)
\( \to a = {m_{F{e_2}{O_3}}} = 0,2.(56.2 + 16.3) = 32{\text{ gam}}\)
\(b = {m_{F{e_2}{{(S{O_4})}_3}}} = 0,2.(56.2 + 96.3) = 80{\text{ gam}}\)
BTKL:
\({m_{dd}} = {m_{F{e_2}{O_3}}} + {m_{dd\;{{\text{H}}_2}S{O_4}}} = 32 + 300 = 332{\text{ gam}}\)
\( \to C{\% _{F{e_2}{{(S{O_4})}_3}}} = \frac{{80}}{{332}}.100\% = 24,1\% \)
