Consider the following reaction occurring in the blast furnace. kg of iron is produced when and are brought together in the furnace. The value of is _______ (nearest integer). [2025]

Question

Consider the following reaction occurring in the blast furnace.

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

$' x'$ kg of iron is produced when $2.32 \times 10^{3} {kg Fe}_{3} O_{4}$ and $2.8 \times 10^{2} kg CO$ are brought together in the furnace. The value of $' x'$ is _______ (nearest integer).

[Given:
${Molar mass of Fe}_{3} O_{4} = 232 {g mol}^{- 1}$
$Molar mass of CO = 28 {g mol}^{- 1}$
$Molar mass of Fe = 56 {g mol}^{- 1}$ ] [2025]

Smart Achievers · Accepted Answer

(420)

Moles of ${Fe}_{3} O_{4} (n_{{Fe}_{3} O_{4}})$ $= \frac{{Given mass of Fe}_{3} O_{4}}{{Molar mass of Fe}_{3} O_{4}}$

$= \frac{2.32 \times 10^{3} kg}{232 {g mol}^{- 1}} = \frac{2.32 \times 10^{3} \times 10^{3} g}{232 {g mol}^{- 1}} = 10000 mol$

Moles of CO $(n_{CO})$ $= \frac{Given mass of CO}{Molar mass of CO}$

$= \frac{2.8 \times 10^{2} kg}{28 {g mol}^{- 1}} = \frac{2.8 \times 10^{2} \times 10^{3} g}{28 {g mol}^{- 1}} = 10000 mol$

${Fe}_{3} O_{4} + 4 CO ⟶ 3 Fe + 4 {CO}_{2}$

As per stoichiometry of the reaction, 10000 mol ${Fe}_{3} O_{4}$ needs 40000 mol of CO for complete reaction. But only 10000 mol CO is present. So CO is the limiting reagent.
So amount of iron formed depends upon the amount of CO.

By reaction stoichiometry:

${Fe}_{3} O_{4} + 4 CO ⟶ 3 Fe + 4 {CO}_{2}$

$\frac{n_{Fe}}{3} = \frac{n_{CO}}{4}$

$\frac{n_{Fe}}{3} = \frac{10000}{4}$

$n_{Fe} = 3 \times \frac{10000}{4} = 7500 mol$

Mass of Fe $= {}^{n}{Fe} \times Molar mass of Fe = 7500 \times 56 g = 420000 g = 420 kg$

Solution

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