JEE Main 2022 — Some Basic Concepts Of Chemistry Question with Solution

Given,

Mass of $$F{e}_3{O}_4$$ = 4.640 kg = 4640 gm

Molar mass of $$F{e}_3{O}_4$$ = 56 $$\times$$ 3 + 16 $$\times$$ 4 = 232 g

$$\therefore$$ Moles of $$F{e}_3{O}_4=\frac{4640}{232}=20$$

Also, given

Mass of CO = 2.520 kg = 2520 gm

Molar mass of CO = 12 + 16 = 28 gm

$$\therefore$$ Molar of $$CO=\frac{2520}{28}=90$$

Here Fe₃O₄ is limiting reagent as to find limiting reagent, divide the given moles of reactants with their respective stoichiometric coefficient and reactant for which this ratio is minimum will be limiting reagent

For $$F{e}_3{O}_4,\frac{moles}{stoichiometriccoefficient}=\frac{20}{1}$$

For $$CO,\frac{moles}{stoichiometriccoefficient}=\frac{90}{4}=22.5$$

$$\therefore$$ Fe₃O₄ is limiting reagent.

Now produced Fe = 20 $$\times$$ 3 = 60 mol

$$\therefore$$ Weight of Fe = 60 $$\times$$ 56 = 3360 g