Consider the following reaction occurring in the blast furnace:
Fe₃ O₄₍ₛ₎+4 CO_((g)) → 3 Fe₍ₗ₎+4 CO_(2( g))
' x ' kg of iron is produced when 2.32 × 10³ kg Fe₃ O₄ and 2.8 × 10² kg CO are brought together in the furnace. The value of ' x ' is _________ . (nearest integer)
Given: molar mass of Fe₃ O₄=232 g mol⁻¹
molar mass of CO=28 g mol⁻¹
molar mass of Fe=56 g mol⁻¹

Solution

Chemistry ma’am · 4 steps

Equal moles of both reactants, but CO is needed four times over, so CO runs out first and decides the iron.

1. Convert to moles. Work in grams. Fe₃O₄: (2.32 × 10⁶ g)/(232 g mol⁻¹) = 1 × 10⁴ mol. CO: (2.8 × 10⁵ g)/(28 g mol⁻¹) = 1 × 10⁴ mol.
2. Find limiting reagent. Divide each mole value by its coefficient. Fe₃O₄ gives 10⁴/1 = 10⁴ and CO gives 10⁴/4 = 2500. The smaller value belongs to CO, so CO is the limiting reagent. Using all of the Fe₃O₄ would need 4 × 10⁴ mol of CO, and there is only 10⁴ mol.

   n_(Fe) = 3/4 × n_(CO) = 3/4 × 1 × 10⁴ = 7500 mol

3. Mass of iron. 7500 mol × 56 g mol⁻¹ = 4.2 × 10⁵ g = 420 kg.
4. So. The iron produced is 420 kg, so x = 420.

The common slip: 1680. Taking Fe₃O₄ as limiting because its mass is larger gives 3 × 10⁴ mol Fe, which is 1680 kg. Compare moles divided by coefficients, not masses.

The idea to keep. To find the limiting reagent, divide each reactant's moles by its coefficient. The smallest value runs out first, and every product amount comes from that reactant.

Answer from the NTA final answer key. Solution by PrepWiser.