\(PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
\(a.\\ n_{Fe}=\frac{25,2}{56}=0,45\left(mol\right)\\ \rightarrow m_{Fe_3O_4}=\frac{0,45}{3}.232=34,8\left(g\right)\)
\(b.\\ n_{Fe}=0,45\left(mol\right)\\ n_{O_2}=\frac{6,4}{32}=0,2\left(mol\right)\\ TL:\frac{0,45}{3}>\frac{0,2}{2}\rightarrow Fe.du\\ \rightarrow m_{Fe_3O_4}=\frac{0,2}{2}.232=23,2\left(g\right)\)