NJC 2023 Solutions to The Mole Concept _ Stoichiometry Discussion Qn

SH1 H2 Chemistry 2023 The Mole Concept and Stoichiometry Discussion Questions Solutions Calculation of Relative Molecular Mass 1 Bromine consists of two isotopes, 79Br and 81Br, in the relative abundance ratio of 1:1. Bromine exists as diatomic molecule, Br2, at room temperature and pressure. Calculate the possible relative molecular masses, M r, of Br 2 molecules formed by these two isotopes and their relative abundance ratio. [1:2:1] Molecular Formula 79Br79Br 79Br81Br or 81Br79Br 81Br81Br Mr 158 160 162 Relative Ratio 4 1 2 1 2 1 =     2 2 1 2 1 2 1 =     4 1 2 1 2 1 =     Abundance Ratio 1 2 1 Calculation involving Parts Per Million 2 Gardeners sometimes fumigate their greenhouses to rid them of pests and moulds by burning a sulfur ‘candle’. A gaseous concentration of sulfur dioxide of 50 ppm (parts per million) by volume is effective. Calculate how many grams of sulfur a gardener need s to burn in order to produce a concentration of 50 ppm of SO2 in a greenhouse that measures 2 m × 3 m × 4 m. Assume room conditions. [1.61 g] Equation for reaction: S(s) + O2(g) → SO2(g) Volume of greenhouse = (2)(3)(4) = 24 m3 = 24000 dm3 [Recall: 1 m3 = 103 dm3] Given that concentration of SO2 = 50 ppm  In 106 dm3 of air, there is 50 dm3 of SO2,  In 24000 dm3 of air, there is 𝟓𝟎 𝟏𝟎𝟔 x 24000 = 1.20 dm3 of SO2 Amt of SO2 = 1.20 ÷ 24.0 = 0.0500 mol [Recall room conditions = r.t.p = 24.0 dm3 mol-1] Mass of S = 0.0500 x 32.1 = 1.605 = 1.61 g ( to 3 sf) Calculation using Mole Concept 3 Phosgene, COCl2, was once used as a war gas. It is poisonous because when inhaled, it reacts with water in the lungs to produce carbon dioxide gas and hydrochloric acid which causes severe lung damage, leading to death ultimately. (a) Calculate the percentage by mass of chlorine in COCl2. [71.7%] ( ) ( ) 2 2 35.5% by mass of C in COC 100% 12.0 16.0 2 35.5 =  = ++ 71.7 %ll (b) Write a balanced equation for the reaction between COCl2 and H2O. COCl2 + H2O → CO2 + 2HCl 𝒙 𝟐𝟒𝟎𝟎𝟎 𝐱 𝟏𝟎𝟔 = 𝟓𝟎 𝒙 = 1.20 dm3

SH1 H2 Chemistry 2023 (c) Calculate the amount of HCl that will be produced by the complete reaction of 0.430 mol of COCl2. [0.860] Since COCl2 : HCl= 1 : 2, l =  =amount of HC produced 0.430 0.860 mol2 (d) Identify the limiting reagent when 0.200 mol of COCl2 is mixed with 6.20 g of H2O. Hence, calculate the amount of HCl produced at the end of the reaction. [0.400] ( ) 2 2 6.20amount of H O used 2 1.0 16.0 amount of COC used 0.200 mol == + = 0.344 mol l 0.200 mol of COCl2 will require 0.200 mol of H2O, Since 0.344 mol of H2O is present (excess), COCl2 is the limiting reagent. Since COCl2 : HCl is 1 : 2, l =  =amount of HC produced 0.200 0.400 mol