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WJEC GCSE Additional Science Teacher’s Notes Ionic compounds in Question 11: a lithium chloride, LiCl 16.5 % lithium, 83.5 % chlorine b potassium oxide, K 2 O 83 % potassium, 17 % oxygen c sodium sulfide, Na 2 S 59 % sodium, 41 % sulfur d magnesium carbonate, MgCO 3 28.6 % magnesium, 14.3 % carbon, 57.1 % oxygen e calcium nitrate, Ca(NO 3 ) 2 24.4 % calcium, 17.1 % nitrogen, 58.5 % oxygen f beryllium oxide, BeO 36 % beryllium, 64 % oxygen g rubidium carbonate, Rb 2 CO 3 73.9 % rubidium, 5.2 % carbon, 20.9 % oxygen h ammonium sulfate, (NH 4 ) 2 SO 4 21.2 % nitrogen, 6.1 % hydrogen, 24.2% sulfur, 48.5 % oxygen 13. Calculate the total mass of reactants and products in the following reactions: a N 2 (g) + 3H 2 (g) → 2NH 3 (g) Reactants: (2 × 14) + 3 × (2 × 1) = 34 Products: 2 × (14 + (3 × 1)) = 34 b 2H 2 (g) + O 2 (g) → 2H 2 O (g) c Reactants: 2 × (2 × 1) + (2 × 16) = 36 Products: 2 × ((2 × 1) + 16) = 36 Mg (s) + H 2 SO 4 (aq) → MgSO 4 (aq) + H 2 O (g) Reactants: 24 + ((2 × 1) + 32 + (4 × 16)) = 122 Products: (24 + 32 + (4 × 16)) + (2 × 1) = 122 d NaOH (aq) + HNO 3 (aq) → NaNO 3 (aq) + H 2 (l) Reactants: (23 + 16 + 1) + (1 + 14 + (3 × 16)) = 103 Products: (23 + 14 + (3 × 16) + ((2 × 1) +16) = 103 14. Calculate the mass of calcium oxide formed by the complete decomposition of 5 kg of calcium carbonate. CaCO 3 (s) → CaO (s) + CO 2 (g) Relative formula mass of calcium carbonate = 40 + 12 + (3 × 16) = 100 Relative formula mass of calcium oxide = 40 + 16 = 56 So 100 g of calcium carbonate makes 56 g of calcium oxide 1 g of calcium carbonate makes (56/100) = 0.56 g of calcium oxide 5 kg = 5000 g of calcium carbonate makes 5000 × 0.56 = 2800 g of calcium oxide = 2.8 kg 15. Calculate the mass of sodium chloride that can be formed from the neutralisation of 8 kg of sodium hydroxide with hydrochloric acid. NaOH (aq) + HCl (aq) → NaCl (aq) + H 2 O (l) Relative formula mass of sodium hydroxide = 23 + 16 + 1 = 40 Relative formula mass of sodium chloride = 23 + 35 = 58 So 40 g of sodium hydroxide makes 58 g of sodium chloride 1 g of sodium hydroxide makes (58/40) = 1.45 g of sodium chloride 8 g of sodium hydroxide makes 8 x 1.45 = 11.6 g of sodium chloride 40
WJEC GCSE Additional Science Teacher’s Notes 16. Calculate the mass of calcium chloride that can be formed by the reaction of 3 g of calcium cabonate with an excess of hydrochloric acid. CaCO 3 (s) + 2HCl (aq) → CaCl 2 (aq) + H 2 O (l) + CO 2 (g) Relative formula mass of calcium carbonate = 40 + 12 + (3 × 16) = 100 Relative formula mass of calcium chloride = 40 + (2 × 35) = 110 So 100 g of calcium carbonate makes 110 g of calcium chloride 1 g of calcium carbonate makes (110/100) = 1.1 g of calcium chloride 3 g of calcium carbonate makes 3 × 1.1 = 3.3 g of calcium chloride PRACTICAL Calculating the formula of copper(II) oxide (pages 124–25) Safety: make sure you consult the relevant CLEAPSS guidance before conducting this practical work. Answers to Questions 1–5 will depend on the Students’ results. The ratio should be approximately 1:1. PRACTICAL Calculating the formula of magnesium oxide (pages 125–26) Safety: make sure you consult the relevant CLEAPSS guidance before conducting this practical work. Answers to Questions 1–5 will depend on the Students’ results. The ratio should be approximately 1:1. Questions 17. Calculate the percentage yield for the Practical Calculating the formula of magnesium oxide (pages 125–6). Answers will depend on Students’ results from the Practical. 18. Six tonnes of ethanol is produced from 15 tonnes of ethene when it is reacted with an excess of water. Calculate the percentage yield of the reaction. C 2 H 4 (g) + H 2 O (g) → C 2 H 5 OH (g) Relative molecular mass of ethene = (2 × 12) + (4 × 1) = 28 Relative molecular mass of ethanol = (2 × 12) + (6 × 1) + 16 = 46 So 28 tonnes of ethene produces 46 tonnes of ethanol 1 tonne of ethene produces (46/28) = 1.64 tonnes of ethanol 15 tonnes of ethene produces (15 × 1.64) = 24.6 tonnes of ethanol Percentage yield 6 100 24.4 % 24.6 19. Find the percentage yield of the reaction if 5 g of CO 2 is realised from the decomposition of 10 g of CaCO 3 . CaCO 3 (s) → CaO (s) + CO 2 (g) 41
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