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58 ml of N/5 H2S04 are used to neutralize ammonia given by 1 gram of organic compound .percentage of Nitrogen in the compound is 34.3 82.7 16.2 21.6?
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58 ml of N/5 H2S04 are used to neutralize ammonia given by 1 gram of o...
Calculation of Percentage of Nitrogen in Organic Compound


  • Given volume of N/5 H2S04 used = 58 ml

  • Let x be the number of moles of N in the organic compound

  • Number of moles of H2S04 used = (N/5) x 0.058

  • From the balanced chemical equation, 1 mole of H2S04 reacts with 2 moles of NH3

  • Therefore, number of moles of NH3 produced = 2 x (N/5) x 0.058

  • Mass of NH3 produced = 2 x (N/5) x 0.058 x 17 (molar mass of NH3)

  • Given mass of organic compound = 1 gram

  • Therefore, percentage of N in organic compound = (mass of N/mass of organic compound) x 100

  • mass of N = percentage of N x mass of organic compound/100

  • Equating the moles of N in organic compound and NH3 produced,

  • x = 2 x (N/5) x 0.058 x 14 (molar mass of N)

  • Substituting for x and mass of NH3 in the above equation,

  • mass of N = (68.8/mass of organic compound) x 100

  • Given percentage of N in organic compound can be calculated by substituting different values of mass of organic compound and checking which gives the correct percentage of N.



Therefore, the percentage of Nitrogen in the organic compound is 16.2%.
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58 ml of N/5 H2S04 are used to neutralize ammonia given by 1 gram of o...
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58 ml of N/5 H2S04 are used to neutralize ammonia given by 1 gram of organic compound .percentage of Nitrogen in the compound is 34.3 82.7 16.2 21.6?
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