\(a.4Fe+3O_2-^{t^o}\rightarrow2Fe_2O_3\\ n_{Fe}=0,3\left(mol\right);n_{Fe_2O_3}=0,1\left(mol\right)\\ TheoPT:n_{Fe\left(pứ\right)}=2n_{Fe_2O_3}=0,2\left(mol\right)< n_{Fe\left(bđ\right)}\\ \Rightarrow Fedư,O_2pứhết\\ b.n_{O_2}=\dfrac{3}{2}n_{Fe_2O_3}=0,15\left(mol\right)\\ \Rightarrow V_{O_2}=0,15.22,4=3,36\left(l\right)\\ m_{Fe\left(dư\right)}=\left(0,3-0,2\right).56=5,6\left(g\right)\)
a.4Fe+3O2−to→2Fe2O3nFe=0,3(mol);nFe2O3=0,1(mol)TheoPT:nFe(pứ)=2nFe2O3=0,2(mol)<nFe(bđ)⇒Fedư,O2pứhếtb.nO