1.(a) (b) 1.3(c) moles of CaCl 2 = 1.23/111.1 = moles Ca 3 (PO 4 ) 2 = 1/3 × = mass Ca 3 (PO 4 ) 2 = × = 1.15 g 3.moles H 2 O 2 = 3.60/34 = mol moles O 2 produced = 1/2 × = mol volume of O 2 (g) = 24 dm 3 mol −1 × mol = 1.27 dm 3 4.volume of Cl 2 needed = 4 × volume CH 4 = 4 × 25.0 cm 3 = 100 cm 3 5.(a) moles of S = mass/molar mass = 32 × 10 6 g/32 g mol −1 = 1 × 10 6 = moles of H 2 SO 4 mass of H 2 SO 4 = moles × molar mass = 1 × 10 6 × 98 = 98 × 10 6 g = 98 tonne % yield = (95/98) x 100 = 99% (b) The atom economy is 100% (the equation has only one substance on the right of the equation) 6.molar mass = (6 × 12) (6 × 16) = 180 g mol −1 mass of carbon in 1 mol = 6 × 12 = 72 g % carbon = (72/180) × 100 = 40% © Hodder & Stoughton Calculations from chemical Answers to equationsTest yourself questions

7.molar mass of NaHCO 3 = 84 g mol −1, so moles taken = mass/molar mass = 4.20/84 = 0.05 mol If equation 1: moles of Na 2 O = ½ × 0.5 = 0.025: mass = moles × molar mass = × 62 = 1.56 g  g, so not equation 1. If equation 2: moles of Na 2 CO 3 = 0.025, so mass = × 106 = 2.65 g, so equation 2 is correct. (If equation 3: moles of NaOH = 0.05, so mass = 0.05 × 40 = 2.0 g, so equation 3 is incorrect) 8.(a) moles = volume × concentration = dm 3 × mol dm −3 = 2.6 × 10 −3 mol (b) volume = moles/concentration = mol/ mol dm −3 = 0.52 dm 3 = 520 cm 3 9.moles of acid = concentration × volume = mol dm −3 × dm 3 = mol moles of NaOH = 2 × = mol concentration of NaOH = moles/volume = mol/ dm 3 = mol dm −3 © Hodder & Stoughton Calculations from chemical Answers to equationsTest yourself questions

10.Error in weighing = ± 2 × 0.01 = ± 0.02 g% error = (0.02/0.45) × 100 = 4.4% Error in pipette = ± 0.05 cm 3 % error = (0.05/10) × 100 = 0.5% Error in titre = ± 2 × 0.05 = ± 0.10 cm 3 % error = (0.10/12.25) × 100 = 0.82% (a) Lowest error is that using the pipette (b) Total error = = 5.7% © Hodder & Stoughton Calculations from chemical Answers to equationsTest yourself questions