Chapter 7 Stoichiometry Worksheet 1 Solution

1 Chapter 7 Mole Concept and Stoichiometry 3E Chapter 7 Mole Concept and Stoichiometry Worksheet One Solutions Modified June 2024 1. In an experiment, 4.0 g of sulfur was burnt in 48.0 dm3 of oxygen measured at r.t.p to form sulfur dioxide. (a) Write the equation for the reaction between sulfur and oxygen. S (s) + O2 (g) → SO2 (g) (b) What was the limiting reactant in this reaction? Number of moles of S given = 4.00 32.0 = 0.125 mol (3 s.f.) Number of moles of O2 given = 48.0 24.0 = 2.00 mol (3 s.f.) Mole ratio of S : O2 = 1 : 1 = 0.125 mol : 0.125 mol (< 2.00 mol given) Since 0.125 mol of S is needed to react with 0.125 mol of O2 but 2 mol of O2 was given, O2 is in excess. Thus, S is the limiting reactant. (c) Calculate the volume of sulfur dioxide formed at r.t.p. Mole ratio of S : O2 = 1 : 1 = 0.125 mol : 0.125 mol Volume of SO2 produced = 0.125 × 24.0 = 3.00 dm3 (3 s.f.) Ans: (b) S (c) 3.00 dm3

2 Chapter 7 Mole Concept and Stoichiometry 2. (a) A compound contains 40.0% carbon, 6.70% hydrogen and 53.3% oxygen. What is its empirical formula? Let the mass of compound be 100 g. Element C H O Mass / g 40 6.7 53.3 Relative atomic mass 12 1 16 Number of moles / mol 40.0 12.0 = 3.33 6.70 1.00 = 6.70 53.3 16.0 = 3.33 Mole ratio 1 2 1 Hence, the empirical formula is CH2O. (b) Given that the compound has an Mr of 180, find its molecular formula. Mr of CH2O = 12 + 1 + 1 + 16 = 30.0 (3 s.f.) n = 180 ÷ 30.0 = 6.00 (3 s.f.) Hence the molecular formula of the compound is C6H12O6. Ans: (a) CH2O (b) C6H12O6

3 Chapter 7 Mole Concept and Stoichiometry 3. In an experiment, 1.20 g of magnesium was reacted with excess hydrochloric acid. Magnesium chloride and hydrogen gas were produced. (a) Write a balanced chemical equation for this reaction. Mg + 2HCl → MgCl2 + H2 (b) Calculate the mass of magnesium chloride produced in this reaction. Number of moles of Mg used = 1.20 ÷ 24 = 0.05 mol Mass of MgCl2 produced = 0.05 × 95 = 4.75 g (c) Calculate the volume of hydrogen gas produced at room temperature and pressure. Number of moles of H2 produced = 0.0500 mol (3 s.f.) Volume of H2 produced = 0.05 24 = 1.20 dm3 (3 s.f.) Ans: (b) 4.75 g (c) 1.20 dm3

4 Chapter 7 Mole Concept and Stoichiometry 4. (a) Define relative atomic mass. The relative atomic mass of an element is the average mass of one atom of that element relative to the mass of an atom of carbon – 12. (b) Define relative molecular mass. The relative molecular mass of a molecular substance is the average mass of one molecule of that substance relative to the mass of an atom of carbon – 12. (c) Calculate the relative molecular mass of the following substances. MgCl2 95 NaOH 40 H2SO4 98 Nitrogen Gas 28 Ans: (Refer to solutions)

5 Chapter 7 Mole Concept and Stoichiometry 5. A magnesium ribbon was loosely coiled and placed in a weighed crucible. The crucible was heated to allow the magnesium to react with oxygen in the air to form magnesium oxide. Mass of Crucible / g : 20.10 Mass of Crucible With Magnesium / g : 20.58 Mass of Crucible With Magnesium Oxide / g : 20.90 (a) Write a balanced chemical equation for this reaction. 2Mg + O2 → 2MgO (b) Calculate the mass of magnesium ribbon used. Mass of Mg = 20.58 – 20.10 = 0.48 g (c) Find the volume of oxygen that reacted. Number of moles of Mg = 0.48 ÷ 24 = 0.0200 mol (3 s.f.) Number of moles of O2 = 0.02 ÷ 2 = 0.0100 mol (3 s.f.) Volume of O2 = 0.01 x 24 = 0.240 dm3 (3 s.f.) Ans: (b) 0.48 g (c) 0.24 dm3