Module 6 Unit 1 Assignment B

MODULE 6 UNIT 1 Assignment Part B By: Connie Warburton 1. 0.673 g of potassium chlorate was decomposed to produce oxygen. The gas was collected over water at 29 o C and 93.33 kPa. Calculate the volume of the collected gas. 2KClO 3 (s) → 2KCl (s) + 3O 2 (g) Pwater = 4.00 kPa Ptotal = 102.66 kPa Vtotal = 0.673g Poxygen = 102.66 kPa – 4.00 kPa = 98.66 kPa Voxygen = 0.673 g x 98.66 kPa ÷ 102.66 kPa Voxygen = 0.646777518 Voxygen = 0.647 g (5 marks) 2. What volume of dinitrogen oxide gas (N 2 O, nitrous oxide or laughing gas) at 18.0°C and 118.16 kPa will form from the complete decomposition of 45.987 g ammonium nitrate, NH 4 NO 3 ? NH 4 NO 3 ( s ) → N 2 O( g ) + 2H 2 O( l ) 1 mol NH 4 NO 3 = (2 x 14.007g) + (4 x 1.008g) + (3 x 16.00g) = 80.046 g 45.987g NH 4 x 1 mol ÷ 80.046g = 0.5745071584 = 0.575 mol NH 4 NO 3 0.575 mol NH 4 NO 3 x 1 mol N 2 O ÷ 1 mol NH 4 OH 3 = 0.575 mol N 2 O 1 mol = 22.4 L 0.575 mol N 2 O x 22.4L ÷ 1 mol = 12.88 L N 2 O Vf = 12.88 L N 2 O x 291K ÷ 273K x 101kPa ÷ 118.16kPa Vf = 11.7353783657 Vf = 11.74 L N 2 O (5 marks) 3. Assume that 13.5 grams of solid aluminum react with HCl according to the following balanced equation at STP : 2 Al (s) + 6 HCl (aq)  2 AlCl 3 (aq) + 3 H 2 (g) How many liters of H 2 are produced? 1 mol Al = 26.982 g 13.5 g Al x 1 mol ÷ 26.982 g = 0.5003335557 = 0.5003 mol Al 0.5003 mol Al x 3 mol H 2 ÷ 2 mol Al = 0.75045 = 0.7505 mol H 2 1 mol = 22.4 L 0.7505 mol H 2 x 22.4 L ÷ 1 mol = 16.8112 = 16.81 L H 2 (5 marks) 1