Đáp án:

a) 49,88% và 50,12%

 b) 5,04l

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
n{H_2} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
nFe = n{H_2} = 0,15\,mol\\
\% mFe = \dfrac{{0,15 \times 56}}{{16,84}} \times 100\%  = 49,88\% \\
\% mF{e_2}{O_3} = 100 - 49,88 = 50,12\% \\
b)\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
nS{O_2} = 0,15 \times 1,5 = 0,225\,mol\\
VS{O_2} = 0,225 \times 22,4 = 5,04l
\end{array}\)

$n_{H_2}=336/22,4=0,15mol$

$PTHH :$

$Fe+H_2SO_4\to FeSO_4+H_2↑(1)$

$Fe_2O_3+3H_2SO_4\overset{t^o}\to Fe_2(SO_4)_3+3H_2O$

 Theo pt (1) :

$n_{Fe}=n_{H_2}=0,15mol⇒m_{Fe}=0,15.56=8,4g$

$⇒\%m_{Fe}=\dfrac{8,4.100\%}{16,84}=49,88\%$

$⇒\%m_{Fe_2O_3}=100\%-49,88\%=50,12\%$

$b/2Fe+6H_2SO_4\overset{t^o}\to Fe_2(SO_4)_3+6H_2O+3SO_2↑$

Theo pt :

$n_{SO_2}=3/2.n_{Fe}=3/2.0,15=0,225mol$

$⇒V_{SO_2}=0,225.22,4=5,04l$