1. Given reaction

$\mathrm{Fe}_3\mathrm{O}_{4(s)} + 4\mathrm{CO}_{(g)} \rightarrow 3\mathrm{Fe}_{(l)} + 4\mathrm{CO}_{2(g)}$

We must find the mass of iron produced from:

• $2.32 \times 10^3\,\text{kg } \mathrm{Fe}_3\mathrm{O}_4$
• $2.8 \times 10^2\,\text{kg } \mathrm{CO}$

2. Convert masses into moles

For $\mathrm{Fe}_3\mathrm{O}_4$

$2.32 \times 10^3\,\text{kg} = 2.32 \times 10^6\,\text{g}$

Moles of $\mathrm{Fe}_3\mathrm{O}_4$:

$n(\mathrm{Fe}_3\mathrm{O}_4) = \frac{2.32 \times 10^6}{232} = 10^4\,\text{mol}$

For CO

$2.8 \times 10^2\,\text{kg} = 2.8 \times 10^5\,\text{g}$

Moles of CO:

$n(\mathrm{CO}) = \frac{2.8 \times 10^5}{28} = 10^4\,\text{mol}$

3. Find the limiting reagent

From the balanced equation:

$1\,\text{mol } \mathrm{Fe}_3\mathrm{O}_4 \text{ requires } 4\,\text{mol CO}$

For $10^4$ mol $\mathrm{Fe}_3\mathrm{O}_4$, required CO would be:

$4 \times 10^4\,\text{mol}$

But available CO is only:

$10^4\,\text{mol}$

So, CO is the limiting reagent.

4. Calculate moles of Fe produced

From the reaction:

$4\,\text{mol CO} \rightarrow 3\,\text{mol Fe}$

Therefore, from $10^4$ mol CO:

$n(\mathrm{Fe}) = \frac{3}{4} \times 10^4 = 7.5 \times 10^3\,\text{mol}$

5. Convert moles of Fe into mass

Molar mass of Fe = $56\,\text{g mol}^{-1}$

$m(\mathrm{Fe}) = 7.5 \times 10^3 \times 56 = 4.2 \times 10^5\,\text{g}$

Convert to kg:

$4.2 \times 10^5\,\text{g} = 420\,\text{kg}$

6. Final answer

$x = 420$

Nearest integer = 420.