..ooo AT&T; 5:50 PM genchem.rutgers.edu MUNTASIR SALEH 2/11/2016 9:34:10 PM Pre-

Question

..ooo AT&T; 5:50 PM genchem.rutgers.edu MUNTASIR SALEH 2/11/2016 9:34:10 PM Pre-Laboratory Assignment for Solvay Process Preparation of Sedium Bicarbonate Before doing the Solvay Process experiment, you must . Correctly solve problems 1, 2, and 3 below With regard to this experiment, be able to Write formulas for the compounds Write and balance equations for the reactions e Pass a short computer quiz (7 correct out of 8 questions) Remember.... NaCI+H20CO2 +NH3NaHCO3+NH4CI 1) You dissolve an excess of NaCl in a solution containing 45.0 mL of 3.53 M ammonia. You bubble in an excess of Co2. Calulate the theoretical yield of sodium bicarbonate. 2)You dissolve 6.87 g of NaCl in a solution containing an excess of ammonia. You bubble in an excess of Co Calulate the theoretical yield of sodium bicarbonate. 3) In problem 1 above, in which you could theoretically convert 45.0 mL of 3.53 M ammonia into sodium bicarbonate, calculate the per cent yield if you actually obtained 7.5086 g of pure sodium bicarbonate Calulated results: Problem 1 a) Calculated moles of ammonia used b) Calculaled theoretical yield of NaliCog Problem 2 e) Calculated moles of NaCI usesd d) Calculated theoretical yield of NaHCo Problem 3: el Calcelaled per cest ied

Explanation / Answer

Problem 1

45 ml of 3.53M Ammonia = 45 X 3.53 /1000 = 0.15885 Moles of NaHCO3 will be produced

Theoritical yield of NaHCO3 = 0.15885 X 84 = 13.34 gm

Problem 2

6.87 gm of NaCl = 6.87 /58.45 = 0.11753 Moles

Theoritical yield of NaHCO3 = 0.11753 X 84 = 9.873 gm

Problem 3

Percentage yield = 7.5086 x 100 /13.34 = 56.29%

Results

Problem 1

0.15885 Moles of ammonia used

Theoritical yield of NaHCO3 = 0.15885 X 84 = 13.34 gm

Problem 2

0.11753 Moles of NaCl used

Theoritical yield of NaHCO3 = 0.11753 X 84 = 9.873 gm

Problem 3

Percentage yield = 7.5086 x 100 /13.34 = 56.29%

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