$PTPƯ:$

$Fe+H_2SO_4\xrightarrow{} FeSO_4+H_2↑$  $(1)$

$Fe_2O_3+3H_2SO_4\xrightarrow{} Fe_2(SO_4)_3+3H_2O$  $(2)$

$n_{H_2}=\dfrac{0,448}{22,4}=0,02mol.$

$Theo$ $pt$ $(1):$ $n_{Fe}=n_{H_2}=0,02mol.$

$⇒\%m_{Fe}=\dfrac{0,02.56}{2}.100\%=56\%$

$⇒\%m_{Fe_2O_3}=100\%-56\%=44\%$

$⇒n_{Fe_2O_3}=\dfrac{2-(0,02.56)}{160}=0,0055mol.$

$Theo$ $pt$ $(1):$ $n_{H_2SO_4}=n_{Fe}=0,02mol.$

$Theo$ $pt$ $(2):$ $n_{H_2SO_4}=3n_{Fe_2O_3}=0,0165mol.$

$⇒n_{H_2SO_4}(2pt)=0,02+0,0165=0,0365mol.$

$\text{Đổi 20 ml = 0,02 lít.}$

$⇒CM_{H_2SO_4}=\dfrac{0,0365}{0,02}=1,825M.$

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Đáp án:

\(\begin{array}{l}
\% {m_{Fe}} = 56\% \\
\% {m_{F{e_2}{O_3}}} = 44\% \\
C{M_{{H_2}S{O_4}}} = 1,825M
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{{H_2}}} = 0,02mol\\
 \to {n_{Fe}} = {n_{{H_2}}} = 0,02mol\\
 \to {m_{Fe}} = 0,02 \times 56 = 1,12g\\
 \to {m_{F{e_2}{O_3}}} = 0,88g \to {n_{F{e_2}{O_3}}} = 0,0055mol\\
 \to \% {m_{Fe}} = \dfrac{{1,12}}{2} \times 100\%  = 56\% \\
 \to \% {m_{F{e_2}{O_3}}} = \dfrac{{0,88}}{2} \times 100\%  = 44\% \\
{n_{{H_2}S{O_4}}} = {n_{Fe}} + 3{n_{F{e_2}{O_3}}} = 0,0365mol\\
 \to C{M_{{H_2}S{O_4}}} = \dfrac{{0,0365}}{{0,02}} = 1,825M
\end{array}\)