0.53g of acetanilide was subjected to kjeldahl determination and the ammonia produced was collected in 50cm3 of 0.50M of h2so4.on dilution 250cm3, the acid was titrated against 25cm3 portion of 0.05M sodium carbonate solution. 13.60 of acid was required for each titration. What is percentage of nitrogen in acetanilide ​

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:

Moles Na₂CO₃ anf Moles H₂SO₄ in excess:

0.025L * (0.05mol / L) = 1.25x10⁻³ moles Na₂CO₃ / 0.01360L =

0.09191M * 0.250L = 0.0230 moles H₂SO₄ in excess.

Moles H₂SO₄ added:

0.050L * (0.50mol / L) = 0.0250 moles H₂SO₄ added

Moles that react:

0.0250 moles - 0.0230 moles = 0.0020 moles H₂SO₄

Moles of NH₃ = Moles N:

0.0020 moles H₂SO₄ * (2mol NH₃ / 1mol H₂SO₄) = 0.0040 moles NH₃ = Moles N

mass N and mass percent:

0.0040 moles N * (14g / mol) = 0.056gN / 0.53g * 100 =

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