Determine the limiting reactant in a reaction and calculate the amount of product formed PPTX.
- 1. Determine the limiting reactant in a reaction and calculate the amount of product formed Group 13
- 2. Step 1: Write the balanced chemical equation. Example: •N₂ + 3H₂ 2NH₃ →
- 3. Step 2: Convert given amounts to moles. • Use the formula: • moles = given mass ÷ molar mass
- 4. Step 3: Use mole ratio (from balanced equation) to compare reactants. Divide the actual moles of each reactant by its coefficient in the balanced equation. •The smallest value indicates the limiting reactant.
- 5. Example Problem: •5.0 grams of N₂ •5.0 grams of H₂ Find the limiting reactant.
- 6. Solution: Balanced equation: •N₂ + 3H₂ 2NH₃ → • Mollar Masses: •N₂ = 28.0 g/mol •H₂ = 2.0 g/mol
- 7. •Convert to moles: •N₂: 5.0 g ÷ 28.0 g/mol = 0.179 mol •H₂: 5.0 g ÷ 2.0 g/mol = 2.5 mol Mole ratio (from balanced equation): •N₂ needs 1 mole •H₂ needs 3 moles
- 8. Now, divide actual moles by required coefficient: N₂: 0.179 ÷ 1 = 0.179 •H₂: 2.5 ÷ 3 = 0.833 Compare: •0.179 (N₂) < 0.833 (H₂) •Therefore, N₂ is the limiting reactant.
- 9. From Before: •Limiting reactant = N₂ •Moles of N₂ = 0.179 mol • •Balanced equation: •N₂ + 3H₂ 2NH₃ →
- 10. Step 1: Use mole ratio to find moles of NH₃ produced. From the balanced equation: •1 mole of N₂ produces 2 moles of NH So: •0.179 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 0.358 mol NH₃
- 11. Step 2: Convert moles of NH₃ to grams. Molar mass of NH₃ = 17.0 g/mol So: •0.358 mol × 17.0 g/mol = 6.09 grams of NH₃
- 12. Final Answer: Approximately 6.1 grams of NH₃ will be formed.