JEE Chemistry 2025 Question with Solution

Given:

• Reaction: $Fe_3O_4(s) + 4CO(g) \rightarrow 3Fe(l) + 4CO_2(g)$
• Mass of $Fe_3O_4 = 2.32 \times 10^3 \, \text{kg} = 2.32 \times 10^6 \, \text{g}$
• Mass of $CO = 2.8 \times 10^2 \, \text{kg} = 2.8 \times 10^5 \, \text{g}$

Find: Mass of iron produced, $x$.

First convert the given masses to moles.

From the balanced equation, $1$ mole of $Fe_3O_4$ reacts with $4$ moles of $CO$. Thus, for $10^4$ moles of $Fe_3O_4$, the required moles of $CO$ are:

But only $10^4$ moles of $CO$ are available, so CO is the limiting reactant and $Fe_3O_4$ is in excess.

Now use the stoichiometric ratio between $CO$ and $Fe$.

Therefore,

Convert moles of iron to mass.

Therefore, the value of $x$ is $420$.

Given: $Fe_3O_4(s) + 4CO(g) \rightarrow 3Fe(l) + 4CO_2(g)$

Find: The mass of iron formed.

Calculate molar masses:

Now find the available moles:

Check the limiting reagent using the stoichiometric ratio $Fe_3O_4 : CO = 1:4$. For $10000$ moles of $Fe_3O_4$, the required $CO$ is:

Since only $10000$ moles of $CO$ are present, $CO$ is the limiting reagent.

From the balanced equation:

So,

Hence, the required numerical answer is $420$.