To determine the amount of ammonium sulfate needed to produce ammonia (NH₃) that completely neutralizes 136.5 grams of hydrochloric acid (HCl), we will follow a systematic approach.
1. Calculate Moles of HCl
- The molar mass of HCl is approximately 36.5 g/mol.
- Moles of HCl = Mass / Molar mass
Moles of HCl = 136.5 g / 36.5 g/mol = 3.74 moles.
2. Neutralization Reaction
- The reaction for the neutralization of ammonia with hydrochloric acid is:
NH₃ + HCl → NH₄Cl.
- From the balanced equation, 1 mole of NH₃ neutralizes 1 mole of HCl.
- Therefore, 3.74 moles of HCl will require 3.74 moles of NH₃.
3. Ammonium Sulfate to Ammonia
- The decomposition of ammonium sulfate ((NH₄)₂SO₄) to produce ammonia is:
(NH₄)₂SO₄ → 2 NH₃ + SO₄²⁻.
- From the reaction, 1 mole of (NH₄)₂SO₄ produces 2 moles of NH₃.
- Thus, to produce 3.74 moles of NH₃, we need:
(3.74 moles NH₃) / 2 = 1.87 moles of (NH₄)₂SO₄.
4. Calculate Mass of Ammonium Sulfate
- The molar mass of ammonium sulfate is approximately 132.14 g/mol.
- Mass of (NH₄)₂SO₄ = Moles × Molar mass
Mass of (NH₄)₂SO₄ = 1.87 moles × 132.14 g/mol = 247.4 grams.
Conclusion
- To completely neutralize 136.5 grams of HCl, you will need approximately 247.4 grams of ammonium sulfate.