\(a,PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(So.sánh:\dfrac{0,2}{3}< \dfrac{0,3}{2}\Rightarrow O_2.dư\)
\(Theo.PTHH:n_{O_2\left(pư\right)}=\dfrac{2}{3}.n_{Fe}=\dfrac{2}{3}.0,2=\dfrac{2}{15}\left(mol\right)\\ \Rightarrow n_{O_2\left(dư\right)}=n_{O_2\left(tổng\right)}-n_{O_2\left(pư\right)}=0,3-\dfrac{2}{15}=\dfrac{1}{6}\left(mol\right)\\ \Rightarrow m_{O_2\left(dư\right)}=n.M=\dfrac{1}{6}.32=\dfrac{16}{3}\left(g\right)\)
\(b,Theo.PTHH:n_{Fe_3O_4}=\dfrac{1}{3}.n_{Fe}=\dfrac{1}{3}.0,2=\dfrac{1}{15}\left(mol\right)\\ \Rightarrow m_{Fe_3O_4}=n.M=\dfrac{1}{15}.232=\dfrac{232}{15}\left(g\right)\)
