Module 5 Unit 2 Assignment B

MODULE 5 UNIT 2 Assignment Part B By: Connie Warburton In each solution, provide all calculations to receive full mark. Mol – mass calculations 1. In the reaction: 2HgO → 2Hg + O 2 How many moles of mercury (II) oxide are needed to produce 125 g of oxygen (3 marks)? 1 mol O 2 = 2 x 16.00g = 32.00g 125 g O 2 x 1 mol ÷ 32.00g = 3.90625 = 3.91 mol O 2 3.91 mol O 2 x 2 mol HgO ÷ 1 mol O 2 = 7.82 mol HgO 2. In the reaction N 2 + 2O 2 → N 2 O 4 If 4.0x10 -3 moles of oxygen reacted, how many grams of N 2 were needed? (3 marks) 4.0 x 10 -3 (0.004) mol O 2 x 1 mol N 2 ÷ 2 mol O 2 = 2.0 x 10 -3 (0.002) mol N 2 Molar mass N 2 = 2 x 14.007g = 28.014 g 2.0 x 10 -3 (0.002) mol N 2 x 28.014g ÷ 1 mol = 56.03 x 10 -3 (0.056028) g N 2 Mass – mass calculations 1 What mass of Fe 2 O 3 must be reacted to generate 324 g of Al 2 O 3 in the reaction: Fe 2 O 3 (s) + 2 Al (s)  2 Fe (s) + Al 2 O 3 (s) (5 marks) 1 mol Al 2 O 3 = (2 x 26.982g) + (3 x 16.00g) = 101.964g 324 g Al 2 O 3 x 1 mol ÷ 101.964g = 3.1775920913 = 3.178 mol Al 2 O 3 3.178 mol Al 2 O 3 x 1 mol Fe 2 O 3 ÷ 1 mol Al 2 O 3 = 3.178 mol Fe 2 O 3 1 mol Fe 2 O 3 = (2 x 55.845g) + (3 x 16.00g) = 159.69g 3.178 mol Fe 2 O 3 x 159.69g ÷ 1 mol = 507.49482 = 507.5 g Fe 2 O 3 1. In the reaction Li 3 N (s) + 3H 2 O (l) → NH 3(g) + 3LiOH (aq) 1

What mass of lithium hydroxide is produced when 0.38g of lithium nitride react? (5 marks) 1 mol Li 3 N = (3 x 6.94g) + 14.007g = 34.827g 0.38g Li 3 N x 1 mol ÷ 34.827g = 0.0109110747 = 0.0109 mol Li 3 N 0.0109 mol Li 3 N x 3 mol LiOH ÷ 1 mol Li 3 N = 0.0327 mol LiOH 1 mol LiOH = 6.94g + 16.00g + 1.008g = 23.948g 0.0327 mol LiOH x 23.948g ÷ 1 mol = 0.7830996 = 0.783g LiOH 2. In the reaction Al 2 (SO 4 ) 3 + 3Ca(OH) 2 →2Al(OH) 3 + 3CaSO 4 a) What mass of aluminum (III) hydroxide are produced if 165.7g of aluminum (III) sulfate react? (5 marks) 1 mol Al 2 (SO 4 ) 3 = (2 x 26.982g) + (3 x 32.06g) + (7 x 16.00g) = 262.144g 165.7g Al 2 (SO 4 ) 3 x 1 mol ÷ 262.144g = 0.6320953369 = 0.632 mol Al 2 (SO 4 ) 3 0.632 mol Al 2 (SO 4 ) 3 x 2 mol Al(OH) 3 ÷ 1 mol Al 2 (SO 4 ) 3 = 1.2642 mol Al(OH) 3 1 mol Al(OH) 3 = 26.982g + (3 x 16.00g) + (3 x 1.008g) = 78.006g 1.2642 mol Al(OH) 3 x 78.006g ÷ 1 mol = 98.6151852 = 98.62g Al(OH) 3 b) How many grams of calcium hydroxide are needed to form 6.35g of calcium sulphate? (5 marks) 1 mol CaSO 4 = 40.078g + 32.06g + (4 x 16.00g) = 136.138g 6.35g CaSO 4 x 1 mol ÷ 136.138g = 0.0466438467 = 0.0466 mol CaSO 4 0.0466 mol CaSO 4 x 3 mol Ca(OH) 2 ÷ 3 mol CaSO 4 = 0.0466 mol Ca(OH) 2 1 mol Ca(OH) 2 = 40.078g + (2 x 16.00g) + (2 x 1.008g) = 74.094g 0.0466 mol Ca(OH) 2 x 74.094g ÷ 1 mol = 3.4527804 = 3.45g Ca(OH) 2 4. 2KMnO 4 + 5H 2 SO 3  K 2 SO 4 + 2MnSO 4 + 2H 2 SO 4 + 3H 2 O Calculate the mass of: a. sulphurous acid, H 2 SO 3 , required to react with 6.32 g of potassium permanganate, KMnO 4 ; (5 marks) 1 mol KMnO 4 = 39.098g + 54.938g + (4 x 16.00) = 158.036g 2

References: Fanshawe College. (2023) “Module 5 Unit 2 Presentation” retrieved from https://olmoodle.ontariolearn.com/mod/scorm/player.php? a=8781¤torg=articulate_rise&scoid=17772&sesskey=SGw7RV8hGW &display=popup&mode=normal Ptable.com (2023) “Periodic Table” retrieved from http://ptable.com/image/periodic-table.svg 4