Final answer:
The quantitative reaction of ammonia with sulphuric acid produces ammonium sulphate. The theoretical and actual yields of the product can determine the percentage yield. The specific quantity of 0.6 moles of urea undergoes a two-step process to eventually form ammonium sulphate when neutralized by sulphuric acid.
Step-by-step explanation:
The reaction between ammonia (NH3) and sulphuric acid (H₂SO₄) yields ammonium sulphate ((NH4)₂SO₄), which is a fertilizer. The balanced chemical equation for this neutralization reaction is:
2 NH3(aq) + H₂SO₄(aq) → (NH4)₂SO₄(aq)
To calculate the theoretical yield of ammonium sulphate from 2.0 kg of sulphuric acid, one must first determine the number of moles of H₂SO₄, then use the stoichiometry of the balanced equation to find the moles of (NH4)₂SO₄ produced. Assuming 100% efficiency in conversion, the theoretical yield in grams can be obtained by multiplying the moles of (NH4)₂SO₄ by its molecular mass. However, in reality, a percentage yield must be calculated if only 2.2 kg of fertilizer was actually produced. Percentage yield is computed using the formula:
Percentage Yield = (Actual Yield / Theoretical Yield) × 100%
It is important to know that the reaction of ammonia with sodium hydroxide to release ammonia, which is then neutralized by sulphuric acid, involves a two-step process. First, urea reacts with sodium hydroxide to releaseammonia, as follows:
CO(NH2)2(s) + 2 NaOH(aq) → 2 NH3(g) + Na2CO3(aq) + H₂O(l)
Following this, the released ammonia is available to react with sulphuric acid to form ammonium sulphate.