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Potassium chromate is produced by the following reaction:

\[ \text{4FeCr}_2\text{O}_4 + 8\text{K}_2\text{CO}_3 + 7\text{O}_2 \rightarrow 8\text{CO}_2 + 2\text{Fe}_2\text{O}_3 + 8\text{K}_2\text{CrO}_4 \]

In one experiment, 169 kg of chromite (FeCr₂O₄), 298 kg of potassium carbonate, and 75.0 kg of oxygen were sealed into a reaction vessel and reacted at high temperatures.

If this reaction was repeated and the new percentage yield was 89.9%, what would the new actual yield be?

Answer :

The new actual yield of potassium chromate would be 263.80 kg when the percentage yield is 89.9%.

Given the balanced chemical reaction: 4FeCr₂O₄ + 8K₂CO₃ + 7O₂ → 8CO₂ + 2Fe₂O₃ + 8K₂CrO₄, the reagents initially used were 169 kg of chromite (FeCr₂O₄), 298 kg of potassium carbonate (K₂CO₃), and 75.0 kg of oxygen (O₂).

  1. Calculate the moles of each reactant to find the limiting reagent.
  2. Calculate the theoretical yield of K₂CrO₄ based on the moles of the limiting reagent.
  3. Apply the percentage yield to find the actual yield.

1. Molecular weights: (FeCr₂O₄),= 223.84 g/mol, K₂CO₃ = 138.205 g/mol, O2 = 32 g/mol

  • Moles of (FeCr₂O₄),: 169 kg × (1000 g/kg) / 223.84 g/mol ≈ 755.1 mol
  • Moles of K₂CO₃ : 298 kg × (1000 g/kg) / 138.205 g/mol ≈ 2156 mol
  • Moles of O₂: 75 kg × (1000 g/kg) / 32 g/mol ≈ 2344 mol

2. Based on the stoichiometry, (FeCr₂O₄),is the limiting reagent:

  • Theoretical yield of K₂CrO₄: 755.1 mol FeCr₂O₄ × (8 mol K₂CrO₄/ 4 mol FeCr₂O) ≈ 1510.2 mol
  • Theoretical mass of K₂CrO₄: 1510.2 mol × 194.19 g/mol ≈ 293,222.4 g or 293.22 kg

3. Actual yield with 89.9% efficiency:

New actual yield = 293.22 kg × 0.899

= 263.80 kg

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