Q1. In a neutralisation experiment, 25 cm3 of dilute sulfuric acid was required to react completely with 40 cm3 of a solution of 0.25 mol dm−3 potassium hydroxide. a. Write a balanced chemical equation for the reaction between dilute sulfuric acid and potassium hydroxide. H2SO4 (aq) + 2 KOH (aq) --------ο K2SO4 (aq) + 2 H2O (l) b. Calculate the number of moles of potassium hydroxide solution used in the reaction. πππππ (π) Formula: πππππππ‘π¦ (π) = ππππ’ππ (π) From above molarity equation, we get πππππ = πππππππ‘π¦ ∗ ππππ’ππ π =π∗π Molarity of potassium hydroxide (M) = 0.25 mol/dm3 Volume of potassium hydroxide (V) = 40 cm3 = 0.040 dm3, (1cm3 = 0.001 dm3) Moles of potassium hydroxide =?? π = 0.25 πππ ∗ 0.040 ππ3 ππ3 π = 0.01 ππππ c. How many moles of dilute sulfuric acid would this number of moles of potassium hydroxide react with? From the balanced chemical equation the mole ratio of sulphuric acid (H2SO4) to potassium hydroxide (KOH) is 1:2 0.01 ππππ πΎππ» ∗ 1 πππ π»2ππ4 2 ππππ πΎππ» = 0.005 ππππ π»2ππ4 d. Calculate the concentration of the dilute sulphuric acid. π= π π Moles of sulfuric acid (n) = 0.005 mols Volume of sulphuric acid (V) = 25 cm3 = 0.025 dm3 π= 0.005 ππππ = 0.2 π 0.025 ππ3 e. Which indicator could have been used to determine when neutralisation had just occurred? Phenolphthalein indicator is used Q2. A type of coal contains 0.5% of sulfur by mass. a. Write an equation for the formation of sulfur dioxide gas when this coal is burned. S (in coal) + O2 (g) -------ο SO2 (g) 32.0 g 32.0 g 64.0 g b. What mass of sulfur is contained in 1500 tonnes of coal? 1500 π‘πππππ ∗ 0.5 = 7.5 π‘πππππ ππ π π’πππ’π 100 b. What mass of sulfur dioxide gas would be formed if 1500 tonnes of coal were burned? Since we have 7.5 tonnes of sulphur in 1500 tonnes of coal, when the 7.5 tonnes of sulphur is burned Mass of SO2 formed: 7.5 π‘πππππ ππ’πππ’π ∗ 106 π 64 π ππ2 ∗ = 15 ∗ 106 π ππ ππ2 1 π‘ππππ 32 π π π’πππ’π c. What volume would this mass of sulfur dioxide gas occupy, measured at room temperature and pressure (rtp)? (Ar: O = 16; S = 32. One mole of a gas occupies 24 dm3 at rtp.) 15 ∗ 106 π ππ ππ2 ∗ 1 ππππ ππ2 24 ππ3 ∗ = 5.6 ∗ 106 ππ3 64 π ππ2 1 ππππ ππ2 Q3. a. Copy and complete the above table by calculating the percentage of nitrogen in the fertilisers sodium nitrate and potassium nitrate. (Ar: H = 1; N = 14; O = 16; Na = 23; K = 39; Ca = 40) Percentage of nitrogen in Sodium nitrate (NaNO3): Molarmass of NaNO3 = 23 + 14 + (3*16) = 23 + 14 + 48 = 85 g/mol Mass of Nitrogen in 1 mole of NaNO3 = 14 g % ππ π ππ ππππ3 = 14 ∗ 100 = 16.47 % 85 Percentage of nitrogen in Potassium nitrate (KNO3): Molarmass of KNO3 = 39 + 14 + (3*16) = 39 + 14 + 48 = 101 g/mol Mass of Nitrogen in 1 mole of KNO3 = 39 g % ππ π ππ πΎππ3 = 39 ∗ 100 = 38.61 % 101 b. Including the data you have just calculated, which of the fertilisers contains: (i) The largest percentage of nitrogen? Ammonia solution (NH3) with 82.4% Nitrogen (ii) The smallest percentage of nitrogen? Sodium nitrate (NaNO3) with 16.5% Nitrogen d. Give the chemical name for the fertiliser that goes by the name Nitram®. Ammonium nitrate (NH4NO3) e. Ammonia can be used directly as a fertiliser but not very commonly. Think of two reasons why ammonia is not often used directly as a fertiliser. i). Very high nitrogen percentage in ammonia ii). Aquatic life is harmed by ammonia even at extreme low concentrations. e. Nitram® fertiliser is manufactured by the reaction of nitric acid with ammonia solution according to the equation: NH3(aq) + HNO3(aq) → NH4NO3(aq) A bag of Nitram® may contain 50 kg of ammonium nitrate. What mass of nitric acid would be required to make it? 103 π 1 ππππ ππ»4ππ3 1 ππππ π»ππ3 63 π π»ππ3 50 πΎπ ππ»4ππ3 ∗ ∗ ∗ ∗ 1 πΎπ 80 π ππ»4ππ3 1 ππππ ππ»4ππ3 1 ππππ π»ππ3 = 39375 π = 39.4 πΎπ Q4. The following diagram shows the three states of matter and how they can be interchanged. a. Name the changes A to E. A = Freezing B = Melting C = Condensation D = Vaporization E = Sublimation b. Name a substance which will undergo change E. Iodine, Camphor …etc c. Name a substance which will undergo changes from solid to liquid to gas between 0 °C and 100 °C. Water d. Describe what happens to the particles of the solid during change E. During the sublimation process, the particles of the solid absorb energy to move fast and overcome the inter-particle attractions to become gas. e. Which of the changes A to E will involve: (i) An input of heat energy? B, D, E (ii) An output of heat energy? A, C, E Q5. The table below shows the melting points, boiling points and densities of substances A to D. a. Which substance is a gas at room temperature? B b. Which substance is a liquid at room temperature? D c. Which substances are solids at room temperature? A and C d. Which substance is most likely to be a metal? A e. Which substance will be a liquid at −260 °C? B f. What is the melting point of the least dense nonmetal? -266 oC g. Which substances are gases at 72 °C? B and D Bby questions: 1.
