\(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: 3Fe + 2O2 --to--> Fe3O4
0,6<--0,4
=> mFe = 0,6.56 = 33,6(g)
\(n_{O_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)
\(0.6......0.4\)
\(m_{Fe}=0.6\cdot56=33.6\left(g\right)\)
\(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
\(PTHH:3Fe+2O_2->Fe_3O_4\)
0,13 0,2 0,4
\(m_{Fe}=0,13.56=7,28\left(g\right)\)