Đáp án:

a) \(\% {m_{Fe}} = 25,9\% ;\% {m_{F{e_2}{O_3}}} = 74,1\% \)

b) \({{\text{m}}_{dd{H_2}S{O_4}}} = 70{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(2Fe + 6{H_2}S{O_4}\xrightarrow{{}}F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\)

\(F{e_2}{O_3} + 3{H_2}S{O_4}\xrightarrow{{}}F{e_2}{(S{O_4})_3} + 3{H_2}O\)

Ta có: 

\({n_{S{O_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol}} \to {{\text{n}}_{Fe}} = \frac{{2{n_{S{O_2}}}}}{3} = 0,1{\text{ mol}} \to {{\text{m}}_{Fe}} = 0,1.56 = 5,6{\text{ gam}} \to {{\text{m}}_{F{e_2}{O_3}}} = 16{\text{ gam}} \to {{\text{n}}_{F{e_2}{O_3}}} = \frac{{16}}{{160}} = 0,1{\text{ mol}}\)

\(\% {m_{Fe}} = \frac{{5,6}}{{21,6}} = 25,9\%  \to \% {m_{F{e_2}{O_3}}} = 74,1\% \)

Ta có:

\({n_{{H_2}S{O_{4{\text{ }}}}{\text{phản ứng}}}} = 3{n_{Fe}} + 3{n_{F{e_3}{O_4}}} = 0,1.3 + 0,1.3 = 0,6{\text{ mol}}\)

Trung hòa axit dư

\({H_2}S{O_4} + 2NaOH\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)

Ta có: 

\({n_{{H_2}S{O_4}}} = \frac{1}{2}{n_{NaOH}} = \frac{{0,2.1}}{2} = 0,1{\text{ mol}} \to {{\text{n}}_{{H_2}S{O_4}{\text{ tham gia}}}} = 0,6 + 0,1 = 0,7{\text{ mol}} \to {{\text{m}}_{dd{H_2}S{O_4}}} = \frac{{0,7.98}}{{98\% }} = 70{\text{ gam}}\)