Fe2O3+3H2SO4→Fe2(SO4)3+3H2O

nFe2O3=24/160=0,15mol

mH2SO4=200.39,2/100=78,4g

nH2SO4=78,4/98=0,8mol

nH2SO4 pứ=0,15.3=0,45mol

→nH2SO4 dư=0,8-0,45=0,35mol

mdd sau pứ=24+200=224g

mFe2(SO4)3=0,15.400=60g

C%Fe2(SO4)3=60/224.100%=26,79%

mH2SO4 dư=0,35.98=34,3g

C%H2SO4 dư=34,3/224.100%=15,31%

Đáp án:

\(\begin{array}{l}
{C_\% }{H_2}S{O_4} \text{ dư } = 15,3125\% \\
{C_\% }F{e_2}{(S{O_4})_3} = 26,79\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
{n_{F{e_2}{O_3}}} = \dfrac{{24}}{{160}} = 0,15\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{200 \times 39,2\% }}{{98}} = 0,8\,mol\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
\dfrac{{{n_{F{e_2}{O_3}}}}}{1} < \dfrac{{{n_{{H_2}S{O_4}}}}}{3} \Rightarrow {H_2}S{O_4} \text{ dư }\\
{n_{{H_2}S{O_4}}} \text{ dư }= 0,8 - 0,15 \times 3 = 0,35\,mol\\
{n_{F{e_2}{{(S{O_4})}_3}}} = {n_{F{e_2}{O_3}}} = 0,15\,mol\\
{m_{{\rm{dd}}spu}} = 24 + 200 = 224g\\
{C_\% }{H_2}S{O_4} \text{ dư } = \dfrac{{0,35 \times 98}}{{224}} \times 100\%  = 15,3125\% \\
{C_\% }F{e_2}{(S{O_4})_3} = \dfrac{{0,15 \times 400}}{{224}} \times 100\%  = 26,79\% 
\end{array}\)