Đáp án:

Giải thích các bước giải:

$\begin{array}{l} {n_{F{e_2}{O_3}}} = 0,01\\ PTHH:\,\,F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,01 \to 0,03\,\,\,\,\,\,\,\,\, \to 0,01\\ {n_{{H_2}S{O_4}}} = 0,03 \to {m_{{H_2}S{O_4}}} = 2,94 \to {m_{{\rm{dd}}{H_2}S{O_4}\,}} = 14,7g\\ {n_{F{e_2}{{(S{O_4})}_3}}} = 0,01 \to {m_{F{e_2}{{(S{O_4})}_3}}} = 4g\\ {m_{{\rm{dd}}\,sau\,pu}} = {m_{F{e_2}{O_3}}} + {m_{_{{\rm{dd}}{H_2}S{O_4}\,}}} = 1,6 + 14,7 = 16,3g\\ C{\% _{F{e_2}{{(S{O_4})}_3}}} = \frac{4}{{16,3}}.100 = 24,54\% \end{array}$

 `a)`

`Fe_2O_3 +3H_2SO_4 → Fe_2(SO_4)_3 + 3H_2O`

`0,01→ 0,03 → 0,005(mol)`

`n_{Fe_2O_3} = {1,6}/{160} = 0,01(mol)`

`mdd_{H_2SO_4} = {0,03.98}/{20%} = 14,7(g)`

`b)`

`C%_{Fe_2(SO_4)_3} = {0,005. 400}/{1,6 + 14,7} .100 = 12,3%`