Production of iron in blast furnace follows the following equation
Fe$$_3$$O$$_4$$(s) + 4CO(g) $$\rightarrow$$ 3Fe(l) + 4CO$$_2$$(g)
when 4.640 kg of Fe$$_3$$O$$_4$$ and 2.520 kg of CO are allowed to react then the amount of iron (in g) produced is:
[Given: Molar Atomic mass (g mol$$^{-1}$$): Fe = 56, O = 16, C = 12]

The balanced equation for the reduction of magnetite by carbon monoxide is given by $$Fe_3O_4(s) + 4CO(g) \rightarrow 3Fe(l) + 4CO_2(g)$$. The molar mass of $$Fe_3O_4$$ is calculated as $$3(56) + 4(16) = 168 + 64 = 232\ \text{g/mol}$$; hence, the number of moles of $$Fe_3O_4$$ in a $$4640\ \text{g}$$ sample is $$\frac{4640}{232} = 20\ \text{mol}$$.

Similarly, the molar mass of carbon monoxide is $$12 + 16 = 28\ \text{g/mol}$$, so the number of moles in $$2520\ \text{g}$$ of CO is $$\frac{2520}{28} = 90\ \text{mol}$$. According to the stoichiometry, each mole of $$Fe_3O_4$$ requires 4 moles of CO, so for $$20\ \text{mol}$$ of $$Fe_3O_4$$ the required amount is $$20 \times 4 = 80\ \text{mol CO}$$. Since $$90\ \text{mol CO}$$ are available, carbon monoxide is in excess and $$Fe_3O_4$$ is the limiting reagent.

Each mole of $$Fe_3O_4$$ yields 3 moles of iron, so $$20\ \text{mol}$$ of $$Fe_3O_4$$ produces $$20 \times 3 = 60\ \text{mol Fe}$$; the mass of this iron is $$60 \times 56 = 3360\ \text{g}$$. Therefore, the mass of iron obtained is 3360 g, corresponding to Option C.