$n_{\mathscr{khí}}=\dfrac{2,24}{22,4}=0,1mol$

$\text{Phương trình phản ứng}$

$Fe+H_2SO_4→FeSO_4+H_2↑$

$\text{Theo phương trình suy ra}$

$n_{H_2}=n_{Fe}=n_{H_2SO_4}'=0,1mol$

$m_{Fe}=0,1.56=5,6g$

$⇒\%m_{Fe}=\dfrac{5,6}{10}.100\%=56\%$

$\%m_{Fe_2O_3}=44\%$

$m_{Fe_2O_3}=4,4g$

$n_{Fe_2O_3}=\dfrac{4,4}{160}=0,0275mol$

$Fe_2O_3+3H_2SO_4→Fe_2(SO_4)_3+3H_2O$

$\text{Theo phương trình suy ra}$

$n_{H_2SO_4}''=3.0,0275=0,0825mol$

$n_{H_2SO_4}=0,1+0,0825=0,1825mol$

$V_{ddH_2SO_4}=\dfrac{0,1825}{2}=0,09125l$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,1mol\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
\% Fe = \dfrac{{5,6}}{{10}} \times 100\%  = 56\% \\
\% F{e_2}{O_3} = 100 - 56 = 44\% \\
b)\\
{m_{F{e_2}{O_3}}} = 10 - 5,6 = 4,4g\\
{n_{F{e_2}{O_3}}} = \dfrac{{4,4}}{{160}} = 0,0275mol\\
{n_{{H_2}S{O_4}}} = {n_{Fe}} + 3{n_{F{e_2}{O_3}}} = 0,1825ml\\
{V_{{H_2}S{O_4}}} = \dfrac{{0,1825}}{2} = 0,09125l = 91,25ml
\end{array}\)