Đáp án:

 a) 72,41%

b) 0,64g

c) 6,32 g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
\% mFe = \dfrac{{56 \times 3}}{{56 \times 3 + 16 \times 4}} \times 100\%  = 72,41\% \\
b)\\
3Fe + 2{O_2} \to F{e_3}{O_4}\\
nF{e_3}{O_4} = \dfrac{m}{M} = \dfrac{{2,32}}{{232}} = 0,01\,mol\\
n{O_2} = 2nF{e_3}{O_4} = 0,02\,mol\\
m{O_2} = n \times M = 0,02 \times 32 = 0,64g\\
c)\\
2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
nKMn{O_4} = 2n{O_2} = 0,04\,mol\\
mKMn{O_4} = n \times M = 0,04 \times 158 = 6,32g
\end{array}\)

$a/\%m_{Fe}=\dfrac{56}{232}.100\%=24,14\%$

$b/n_{Fe_3O_4}=2,32/232=0,01mol$

$PTHH : 3Fe+2O_2\overset{t^o}\to Fe_3O_4$

Theo pt :

$n_{O_2}=2.n_{Fe_3O_4}=2.0,01=0,02mol$

$⇒m_{O_2}=0,02.32=0,64g$

$c/2KMnO_4\overset{t^o}\to K_2MnO_4+MnO_2+O_2$

Theo pt :

$n_{KMnO_4}=2.n_{O_2}=2.0,02=0,04mol$

$⇒m_{KMnO_4}=0,04.158=6,32g$