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STUDY BLOCK 3 ANSWERS 1. Balance the equations: (a) (NH 4 ) 2 S + HgBr 2 → 2NH 4 Br + HgS (b) P 4 O 10 + 6H 2 O → 4H 3 PO 4 (c) 2C 4 H 10 + 13O 2 → 8CO 2 + 10H 2 O (d) 2H 2 O 2 → 2H 2 O + O 2 (e) NH 4 NO 2 → N 2 + 2H 2 O (f) 3FeCl 2 + 2Na 3 PO 4 → Fe 3 (PO 4 ) 2 + 6NaCl (g) 3CaCO 3 + 2H 3 PO 4 → Ca 3 (PO 4 ) 2 + 3CO 2 + 3H 2 O (h) C 3 H 7 COOH + 5O 2 → 4CO 2 + 4H 2 O 2. One litre of solution contains 0.100 moles of FeCl 3 and 0.100 moles of NH 4 Cl. Calculate the number of moles of Fe 3+ ions, Cl - ions and NH 4+ From the FeCl ions in the 1 L of solution. 3 In 1 mol of FeCl: 3 there is 1 mol of Fe 3+ and 3 mol of Cl – ∴ In 0.1 mol of FeCl 3 there is 0.1 mol of Fe 3+ and 0.3 mol of Cl – From the NH 4 Cl: In 1 mol of NH 4 Cl there is 1 mol of NH 4+ and 1 mol of Cl ∴ In 0.1 mol of NH – 4 Cl there is 0.1 mol of NH 4+ and 0.1 mol of Cl – Thus in the 1 L solution: n(Fe 3+ n(Cl) = 0.1 mol – n(NH) = 0.3 + 0.1 = 0.4 mol 4+ 3. Classify the following reactions as one of: redox, precipitation, decomposition, acid-base or combustion. (a) Pb) = 0.1 mol 2+ (aq) + 2I - (aq) →PbI 2 (s) Precipitation (b) HCN + OH - → H 2 O + CN - acid-base (c) CaCO 3 → CaO + CO 2 decomposition (d) Mg(s) + Cl 2 (g) →MgCl 2 (s) redox (e) C 2 H 4 + 4O 2 → 2CO 2 + 4H 2 4. How many atoms are there in: (a) 1 mole of carbon atoms, C? 6.02 × 10O + energy combustion 23 C atoms (b) 1 mole of hydrogen atoms, H? 6.02 × 10 23 H atoms (c) 1 mole of hydrogen gas, H 2 ? 2 × 6.02 × 10 23 = 1.204 × 10 24 H atoms 5. Determine the weight in grams of one silver atom. Mass of one mole of Ag atoms = 107.9 g mol i.e. M(Ag) = 107.9 g mol –1 –1 g1079.1 1002.6 9.107 N9.107 atomAgoneof Mass 2223A − ×=×== 6. Calculate the molar mass (molecular weight) of each of the following: (a) SiF 4 M(SiF 4 ) = 104.09 g mol –1 (b) HF M(HF) = 20.008 g mol –1 (c) Cl 2 M(Cl 2 ) = 70.90 g mol –1 (d) Xe M(Xe) = 131.3 g mol –1 (e) NO 2 M(NO 2 ) = 46.01 g mol –1 (f) PtCl 4 M(PtCl 4 ) = 336.90 g mol –1 (g) ZnBr 2 M(ZnBr 2 ) = 225.19 g mol –1 (h) CH 3 NH 2 M(CH 3 NH 2 ) = 31.05 g mol7. How many grams of H –1 2 are there in 1.5 moles of H 2 M(H gas? 2 ) = 2.016 g mol –1 So, m(H 2 ) = n × M(H 2 8. A chemist weighs out 10 g of water (H) = 1.5 × 2.016 = 3.024 = 3.0 g (1sf) 2 O), 10 g of ammonia (NH 3 Water: M(H) and 10 g of hydrogen chloride (HCl). How many moles of each substance does this represent? 2 O) = 18.016 g mol –1 (2sf)mol56.0 016.1810Mm O)n(H 2 === Ammonia: M(NH 3 ) = 17.034 g mol –1 mol59.0 034.1710)n(NH 3 == Hydrogen chloride: M(HCl) = 36.458 g mol –1 mol27.0 36.45810n(HCl) == 9. How many moles of atoms are present in 9.00 g of Al? Calculate the number of aluminium atoms this represents? mol334.0 26.989n(Al) == Number of Al atoms = n × N A = 0.334 × 6.02 ×10 23 = 2.01 × 1010. What is the molecular formula of a compound whose molecular mass is 198 and which contains 24.2% carbon, 4.0% hydrogen, 71.8% chlorine? 23 C H Cl 24.2 4 71.8 015.201.122.24 = 97.3008.14 = 03.245.358.71 = 103.2015.2 ≅ 203.297.3 ≅ 103.203.2 = The empirical formulae is CH 2 Empirical mass = 12.01 + (2 × 1.008) + 35.45 Cl = 49.476 g 449.476198massemprical massMolar == Therefore molecular formula is C 4 H 8 Cl11. A 100 mg sample of a compound containing only C, H and O was found, by analysis, to give 149 mg CO 4 2 and 45.5 mg of H 2 To solve this problem, you need to recognize that all the C in the COO when burned completely. Calculate the empirical formula. 2 srcinated from the unknown compound. Thus if we calculate the mass of C in the CO 2 Similarly, all the H in the H this will give the mass of C in the unknown compound. 2 O srcinated from the unknown compound. Thus if we calculate the mass of H in the H 2 The mass of O can then be found by adding the masses of C and H and subtracting them from the total mass of the unknown sample. O this will give the mass of H in the unknown compound. Mass of C: mol1039.3 44.010.149)()m(CO)n(COn(C) 3222 − ×==== CO M So, m(C) = n × M = 3.39 ×10 -3 × 12.01 = 0.0407 g Mass of H: mol1005.5 18.0160.04552)()Om(H2)On(H2n(H) 3222 − ×=×=×=×= O H M So, m(H) = n × M = 5.05 ×10 -3 × 1.008 = 5.09 × 10 -3 Mass of O: m(O) = 0.1 – (0.0407 + 5.09 × 10 g -3 ) = 0.0542 g C H O 0.0407 0.00509 0.0542 (Divide by molar masses) 0.00339 0.00505 0.00339 (Divide by smallest number) 1 1.49 1 (Round and multiply by 2) 2 3 2 The empirical formula is C 2 H 3 O 2
