Module 5 Unit 2 Assignment C

MODULE 5 UNIT 2 Assignment Part C By: Connie Warburton. This assignment is to be sent to your instructor for evaluation. Show all calculations, molar ratios, molar masses, etc. 1. In the reaction LiOH + KCI ➔ LiCI + KOH, 20 grams of lithium hydroxide, LiOH, was used to produce lithium chloride, LiCl. Calculate the theoretical yield of lithium chloride. (4 marks) What is the percent yield if only 6 grams of LiCl was produced? (2 marks) 1 mol LiOH = 6.94g + 16.00g + 1.008g = 23.948g 20g LiOH x 1 mol ÷ 23.948g = 0.690894016 = 0.691 mol LiOH 0.691 mol LiOH x 1 mol LiCl ÷ 1 mol LiOH = 0.691 mol LiCl 1 mol LiCl = 6.94g + 35.45g = 42.39g 0.691 mol LiCl x 42.39g ÷ 1 mol = 29.291949 = 29.3g LiCl Actual yield ÷ theoretical yield x 100% = percent yield 6g ÷ 29.3g x 100% = 20.4778156997 = 20.5% 2. In the reaction, FeBr 2 + 2 KCI ➔ FeCl 2 + 2 KBr, 34 grams of iron (II) bromide was used. What is the theoretical yield of iron (II) chloride, FeCl 2 ? (4 marks) If 4g of iron (II) chloride was obtained, what is the percent yield? (2 marks) 1 mol FeBr 2 = 55.845g + (2 x 79.904g) = 215.653g 34g FeBr 2 x 1 mol ÷ 215.653g = 0.1622977654 = 0.163 mol FeBr 2 0.163 mol FeBr 2 x 1 mol FeCl 2 ÷ 1 mol FeBr 2 = 0.163 mol FeCl 2 1 mol FeCl 2 = 55.845g + (2 x 35.45g) = 126.745g 0.163 mol FeCl 2 x 126.745g ÷ 1 mol = 20.659435 = 20.7g FeCl 2 Actual yield ÷ theoretical yield x 100% = percent yield 4g ÷ 20.7g x 100% = 19.3236714976 = 19.3% 3. Sn 3 (PO 4 ) 4 + 6Na 2 CO 3 →3Sn(CO3) 2 + 4Na 3 PO 4 If 7.3grams of sodium carbonate, Na 2 CO 3 , are used in the reaction, what is the theoretical yield of sodium phosphate, Na 3 PO 4 ? (4 marks) What is the percent yield if actually, 5.6 g Na 2 PO 4 was formed? (2 marks) 1 mol Na 2 CO 3 = (2 x 22.990g) + 12.011g + (3 x 16.00g) = 105.991g 7.3g Na 2 CO 3 x 1 mol ÷ 105.991g = 0.0688737723 = 0.069 mol Na 2 CO 3 0.069 mol Na 2 CO 3 x 4 mol Na 3 PO 4 ÷ 6 mol Na 2 CO 3 = 0.046 mol Na 3 PO 4 1

1 mol Na 3 PO 4 = (3 x 22.990g) + 30.974g + (4 x 16.00g) = 163.944g 0.046 mol Na 3 PO 4 x 163.944g ÷ 1 mol = 7.541424 = 7.5 g Na 3 PO 4 Actual yield ÷ theoretical yield x 100% = percent yield 5.6g ÷ 7.5g x 100% = 74.6666666667 = 74.7% Total 18 points Submission Instructions Submit your assignment using the drop box on the Assignment page for this unit. Remember to include references. 2