Converting quantities between different substances using a balanced equation The balancing (stoichiometric) numbers are mole ratios e.g. 1 mol of N2 reacts with 3 mol of H2 to produce 2mol of NH3 Typically we are given a quantity of one substance and are asked to work out a quantity for another substance in the reaction. Any of the above three equations can be used. Step 1: Use one of the above 3 equations to convert any given quantity into amount in mol Mass amount Volume of gas amount Conc and vol of solution amount Step 2: Use balanced equation to convert amount in mol of initial substance into amount in mol of second substance Step 3 Convert amount, in mol, of second substance into quantity question asked for using relevant equation e.g. amount ,Mr mass Amount gas vol gas amount, vol solution conc Example 14: What mass of Copper would react completely with 150 cm3 of 1.60M nitric acid? 3Cu + 8HNO3 3Cu(NO3 )2 + 2NO + 4H2O Step 1: work out amount, in mol, of nitric acid amount = conc x vol = 1.60 x 0.15 = 0.24 mol Step 2: use balanced equation to give moles of Cu 8 moles HNO3 : 3 moles Cu So 0.24 HNO3 : 0.09 (0.24 x 3/8 ) mol Cu Step 3: work out mass of Cu Mass = amount x Mr = 0.09 x 63.5 =5.71g Example 12: 23.6cm3 of H2SO4 neutralised 25.0cm3 of 0.150M NaOH. What is the concentration of the H2SO4? H2SO4 + 2NaOH Na2SO4 +2H2O Step 1: work out amount, in mol, of sodium hydroxide amount = conc x vol = 0.150 x 0.025 = 0. 00375 mol Step 2: use balanced equation to give moles of H2SO4 2 moles NaOH : 1 moles H2SO4 So 0.00375 NaOH : 0.001875 mol H2SO4 Step 3 work out concentration of H2SO4 conc= amount/Volume = 0.001875 / 0.0236 = 0.0794 mol dm-3 Example 11: What mass of Carbon dioxide would be produced from heating 5.50 g of sodium hydrogencarbonate? 2NaHCO3 Na2CO3 + CO2 + H2O Step 1: work out amount, in mol, of sodium hydrogencarbonate amount = mass / Mr = 5.5 /84 = 0.0655 mol Step 2: use balanced equation to give amount in mol of CO2 2 moles NaHCO3 : 1 moles CO2 So 0.0655 HNO3 : 0.0328mol CO2 Step 3: work out mass of CO2 Mass = amount x Mr = 0.0328 x 44.0 =1.44g N Goalby chemrevise.org 6 Example 13: What volume in cm3 of oxygen gas would be produced from the decomposition of 0.532 g of potassium chlorate(V)? 2KClO3 2 KCl + 3O2 Step 1: work out amount, in mol, of potassium chlorate(V)? amount = mass / Mr = 0.532 /122.6 = 0.00434 mol Step 3: work out volume of O2 Gas Volume (dm3 )= amount x 24 = 0.00651 x 24 = 0.156 dm3 = 156 cm3 Step 2: use balanced equation to give amount in mol of O2 2 moles KClO3 : 3 moles O2 So 0.00434 HNO3 : 0.00651mol O2

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2.1.3 Amount of substance

Calculation of reacting masses, gas volumes and mole concentrations

(g) use of stoichiometric relationships in calculations

2.1.4 Acids

Acid–base titrations

(e) structured and non-structured titration calculations, based on experimental results of familiar and non-familiar acids and bases.