Find Mass of Iron Using Limiting Reagent JEE Main PYQ

 

❓ Question

Consider the following reaction occurring in a blast furnace:

$$Fe_3O_4(s)+4CO(g)\rightarrow3Fe(l)+4CO_2(g)$$

$x$ kg of iron is produced when $2.32\times10^3$ kg of $Fe_3O_4$ and $2.8\times10^2$ kg of CO are brought together in the furnace. Find the value of $x$.

(Molar masses: $Fe_3O_4=232\text{ g mol}^{-1}$, $CO=28\text{ g mol}^{-1}$, $Fe=56\text{ g mol}^{-1}$)

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✍️ Solution

Moles of $Fe_3O_4$:

$$n=\frac{\text{Given mass}}{\text{Molar mass}}$$
$$n(Fe_3O_4)=\frac{2.32\times10^3\times10^3}{232}$$
$$=10^4\text{ mol}$$

Moles of CO:

$$n(CO)=\frac{2.8\times10^2\times10^3}{28}$$
$$=10^4\text{ mol}$$

From the balanced equation:

$$Fe_3O_4+4CO\rightarrow3Fe+4CO_2$$
$$1\text{ mol }Fe_3O_4\text{ requires }4\text{ mol CO}$$

For $10^4$ mol $Fe_3O_4$, required CO is:

$$4\times10^4\text{ mol}$$

But available CO is only:

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

Hence, CO is the limiting reagent.

From the reaction:

$$4\text{ mol CO}\rightarrow3\text{ mol Fe}$$

Therefore,

$$10^4\text{ mol CO}\rightarrow\frac34\times10^4\text{ mol Fe}$$

Mass of iron produced:

$$m=n\times M$$
$$=\frac34\times10^4\times56$$
$$=4.2\times10^5\text{ g}$$
$$=420\text{ kg}$$

✅ Final Answer

$$\boxed{x=420\text{ kg}}$$