\(n_{Fe_3O_{\text{ 4 }}}=\dfrac{34,8}{232}=0,15\left(mol\right)\\
pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
0,45<-0,3<----------0,15
\(m_{Fe}=0,45.56=25,2\left(g\right)\\
pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
0,6<--------------------------------------0,3
\(m_{KMnO_4}=0,6.158=94,8\left(g\right)\)