JEE Main 2025 Jan 24 Shift 1, Chemistry Q25: Stoichiometry

JEE Main2025Jan 24, Shift 1Chemistry

Consider the following reaction occurring in the blast furnace.

$F e_{3} O_{4} (s) + 4CO (g) \to 3Fe (l) + 4C O_{2} (g)$

' $x$ ' kg of iron is produced when $2.32 \times 1 0^{3}$ kg Fe $_{3}$ O $_{4}$ and $2.8 \times 1 0^{2}$ kg CO are brought together in the furnace. The value of ' $x$ ' is ____. (nearest integer)

{Given :

Molar mass of Fe $_{3}$ O $_{4}$ = 232 g mol $^{- 1}$

Molar mass of CO = 28 g mol $^{- 1}$

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

Your answer

Solution

Step 1 $-$ moles of each reactant (convert kg to g first):

$n (F e_{3} O_{4}) = \frac{2.32 \times 1 0 ^{6} g}{232 g/mol} = 1.0 \times 1 0^{4} mol$

$n (CO) = \frac{2.8 \times 1 0 ^{5} g}{28 g/mol} = 1.0 \times 1 0^{4} mol$

Step 2 $-$ identify limiting reagent. Stoichiometry needs 4 mol CO per 1 mol Fe $_{3}$ O $_{4}$ , i.e. CO requirement $= 4 \times 1 0^{4}$ mol, but only $1.0 \times 1 0^{4}$ mol CO is available. CO is the limiting reagent.

Step 3 $-$ moles of Fe produced. From 4 CO $\to$ 3 Fe:

$n (Fe) = \frac{3}{4} \times n (CO) = \frac{3}{4} \times 1.0 \times 1 0^{4} = 7.5 \times 1 0^{3} mol$

Step 4 $-$ mass of Fe:

$m (Fe) = 7.5 \times 1 0^{3} \times 56 g = 4.2 \times 1 0^{5} g = 420 kg$ .

$∴$ Answer: $420$ .

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JEE Main 2025 Jan 24 Shift 1, Chemistry Q25: Stoichiometry

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