UNIVERSITI TUNKU ABDUL RAHMAN Centre Course Year/ Trimester Lecturer : : : : Centre for Foundation Studies Foundation In Science Year 1 Trimester 1 Unit Code Unit Title Session : : : FHSC1114 Physical Chemistry 2014/05

Ms Azlina Banu, Mr Tam Yun Hong, Ms Chong Pui Kuan, Ms Amelia Chiang Kar Mun, Ms Farhanah, Ms Wong Jing Tyng, Ms Jamie Anne, Ms Lau Mei Chien, Mr Ng Sweet Kin, Ms Phang Ying Ning, Ms Precilla, Ms Rachel Tham, Ms Rajalakshmi, Mr Tan Jun Bin, Ms Tan Lee Siew

Tutorial 3 : Chapter 3 Stoichiometry and Solution Concentration
1. Balance the following equations: (a) Al(s) + NH4ClO4(s) → Al2O3(s) + AlCl3(s) + NO(g) + H2O(g)

(b) GaBr3(aq) + Na2SO3(aq) → Ga2(SO3)3(aq) + NaBr(aq) 2. Ethanol, C2H5OH, is a liquid with a density of 0.789 g ml-1 at 25 °C. Calculate the molarity of ethanol solution made by dissolving 20.00 mL of ethanol at 25 °C in enough water to make 250.0 ml of solution. [Ans: 1.37 mol L-1]

3. Copper sulfate is widely used as a dietary supplement for animal feed. A lab technician prepares a “stock” solution of CuSO4 by dissolving 79.80 g of CuSO4 in enough water to make 500.0 mL of solution. (a) Determine the molarity of the CuSO4 “stock” solution prepared by the technician. [Ans: 1 mol L-1] (b) Calculate the volume of CuSO4 “stock” solution that should be diluted to give 2.5 L of [Ans: 0.25 L] 0.1 M CuSO4. 4. Aluminum is a limiting reactant in the reaction with sulfur gas to form aluminum sulfide. Initially, 1.18 mol of aluminum and 2.25 mol of sulfur are combined. 2Al(s) + 3S(s) → Al2S3(s) (a) Calculate the aluminum sulfide formed in moles. (b) Calculate the mass of the excess reactant after the reaction. [Ans: 0.59 mol] [Ans: 15.39 g] [December 2013]

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5. A mixture of excess carbon disulphide, CS2, and 650 cm3 of 9.2 g dm-3 chlorine, Cl2, are passed through a hot reaction tube to form tetrachloromethane, CCl4, as below: CS2(l) + 3Cl2(g) → CCl4(l) + S2Cl2(l) Determine the mass of tetrachloromethane formed if the percentage yield is 78%. [September 2011] [Ans: 3.37 g] 6. An impure sample of 1.034 g oxalic acid (H2C2O4) is dissolved in water and an acid-base indicator is added. In titration, the sample requires 34.47 mL of 0.485 M NaOH to reach the equivalence point. The equation is shown below: H2C2O4(aq) + 2NaOH(aq) → Na2C2O4(aq) + 2H2O(ℓ) Determine the mass percent of oxalic acid in the sample. [Ans: 72.71%]

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