Consider the following reaction occurring in the blast furnace: $$Fe_3O_4(s) + 4CO(g) \rightarrow 3Fe(l) + 4CO_2(g) x$$ kg of iron is produced when  $$2.32\times10^3\,kg\,Fe_3O_4$$ and $$2.8\times10^2\,kg\,CO$$ are brought together in the furnace. The value of  $$x$$  is  $$\underline{\hspace{2cm}}$$ (nearest integer). Given: $$M(Fe_3O_4)=232\,g\,mol^{-1},$$ molar mass of  $$CO=28\,g\,mol^{-1},$$ molar mass of $$(Fe)=56\,g\,mol^{-1}.$$

The reaction: $$Fe_3O_4(s) + 4CO(g) \to 3Fe(l) + 4CO_2(g)$$

Given: $$2.32 \times 10^3$$ kg $$Fe_3O_4$$ and $$2.8 \times 10^2$$ kg CO.

Moles of $$Fe_3O_4 = \frac{2.32 \times 10^6 \text{ g}}{232 \text{ g/mol}} = 10^4$$ mol.

Moles of CO $$= \frac{2.8 \times 10^5 \text{ g}}{28 \text{ g/mol}} = 10^4$$ mol.

From stoichiometry: 1 mol $$Fe_3O_4$$ needs 4 mol CO.

$$10^4$$ mol $$Fe_3O_4$$ needs $$4 \times 10^4$$ mol CO, but only $$10^4$$ mol CO is available.

CO is the limiting reagent.

From 4 mol CO, we get 3 mol Fe.
From $$10^4$$ mol CO: Fe produced $$= \frac{3}{4} \times 10^4 = 7500$$ mol.

Mass of Fe $$= 7500 \times 56 = 420000$$ g $$= 420$$ kg.

The answer is 420.