Answer:
10.57% of N in acetanilide
Explanation:
All nitrogen in the sample is converted in NH₃ in the Kjeldahl determination. The NH₃ reacts with H₂SO₄ as follows:
2NH₃ + H₂SO₄ → 2NH₄⁺ + SO₄²⁻
The acid in excess in titrated with Na₂CO₃ as follows:
Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂
To solve this question we must find the moles of sodium carbonate = Moles of H₂SO₄ in excess. The added moles - Moles in excess = Moles of sulfuric acid that reacts:
<em>Moles Na₂CO₃ anf Moles H₂SO₄ in excess:</em>
0.025L * (0.05mol / L) = 1.25x10⁻³ moles Na₂CO₃ / 0.01360L =
0.09191M * 0.250L = 0.0230 moles H₂SO₄ in excess.
<em>Moles H₂SO₄ added:</em>
0.050L * (0.50mol / L) = 0.0250 moles H₂SO₄ added
<em>Moles that react:</em>
0.0250 moles - 0.0230 moles = 0.0020 moles H₂SO₄
<em>Moles of NH₃ = Moles N:</em>
0.0020 moles H₂SO₄ * (2mol NH₃ / 1mol H₂SO₄) = 0.0040 moles NH₃ = Moles N
<em>mass N and mass percent:</em>
0.0040 moles N * (14g / mol) = 0.056gN / 0.53g * 100 =
<h3>10.57% of N in acetanilide</h3>
