\(a) 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ b)n_{Fe} = \dfrac{5,6}{56} = 0,1(mol)\\ n_{O_2} = \dfrac{2}{3}n_{Fe} = \dfrac{0,2}{3}(mol)\\ \Rightarrow V_{O_2} = \dfrac{0,2}{3}.22,4 = 1,493(lít)\)
\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
\(0.1....\dfrac{1}{15}\)
\(V_{O_2}=\dfrac{1}{15}\cdot22.4=1.493\left(l\right)\)