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Report on Preparing a sample of fertiliser, in titration

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´╗┐Preparing a sample of fertiliser The Equipment used in this experiment was: 1. Eye Protection. - 25ml Graduated pipette and filler. 2. Burette and Stand. - Litmus indicator. 3. Evaporating Basin. - Tripod. 4. Balance (to 0.0g). - Ammonium Hydroxide. 5. 250ml Conical Flask. - Sulphuric Acid. 6. Clay pipe triangle. - Bunsen Burner. 7. White tile. Before starting this experiment I weighed the Dish, this came to 45.9g. Rough 1 2 3 Average Final Burette Reading (cm3) 22.65 22.15 22.2 22.15 22.16 Initial Burette Reading (cm3) 0 0 0 0 0 Titre (cm3) 22.65 22.15 22.2 22.15 22.16 The calculation to find the average was Final Burette Reading 1 + Final Burette Reading 2 + Final Burette Reading 3 = Average 3 22.5 +22.2 + 22.15 = 22.16 3 Carefully we drew 25ml of Ammonium Hydroxide using a glass pipette. ...read more.


Turning the nozzle slowly on the burette so it was horizontal we allowed the acid to empty into the conical flask with the Ammonia in it, one drop at a time, until the Ammonia solution and Acid, mixed and turned an almost pinkish colour. Recording how much of the Sulphuric Acid was used, we then repeated this 3 times till they were all 0.1cm3 difference between them, we the calculated the average. We repeated the procedure again, this time without the litmus indicator, adding the average volume of acid, as accurately as possible. The reason we did this, was so that there wasn?t any other chemicals mixed into the solution, when we heated the ammonium sulphate later. Pouring the solution, into the pre-weighed evaporating basin. ...read more.


2.85 Solutions mass 22.16 Mass lost 19.31 After several days, we checked our solution, and found that all the liquid had been evaporated, and crystals, were what remained. We measured the crystals, and recorded all the information in the table above. The reaction that had taken place was: Ammonium Hydroxide + Sulphuric Acid Ammonium Sulphate 2NH4OH + H2SO4 (NH4)2SO4 + 2H2O H2SO4 NH4OH (NH4)2SO4 Volume (in dm3) 0.02216 0.025 Concentration 1M 2M Moles = 1 X 0.02216 = 0.02216 mol = 2 X 0.025 = 0.05 mol Ratio (from balanced equation) 1 2 1 Moles = Mass of Crystals Mr (NH4)2 SO4 N X 2 = 14 X 2 = 28 H X 8 = 1 X 8 = 8 S X 1 =32.1 X 1 = 32.1 O X 4 = 16 X 4 = 64 132.1 Moles = 2.85 = 0.02157456 132.1 ...read more.

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