SOLUCIONARIO DE QUÍMICA CUANTITATIVA 2009/10/11
1. D
2. B
3. A
4. B
5. D
6. (a) (i) (10 % 1000 g =) 100 g ethanol and (90 % 1000 g =) 900 g octane; 1
(ii) n(ethanol) = 2.17 mol and n(octane) = 7.88 mol; 1
(iii) Ereleased from ethanol = (2.17 × 1367) = 2966 (kJ); Ereleased from octane = (7.88 × 5470) = 43104 (kJ);
total energy released = (2966 + 43104) = 4.61 × 104 (kJ); 3
Award [3] for correct final answer.
Accept answers using whole numbers for molar masses and rounding.
(b) greater;
fewer intermolecular bonds/forces to break / vaporization is
endothermic / gaseous fuel has greater enthalpy than liquid fuel / OWTTE; 2
M2 cannot be scored if M1 is incorrect.
7. D
8. D
9. D
10. A
11. D
12. C
13. B
14. D
15. A
16. A
17. (a) (i) Copper:
0 to +2 / increases by 2 / +2 / 2+;
Allow zero/nought for 0.
Nitrogen:
+5 to +4 / decreases by 1 / –1 / 1–;
Penalize missing + sign or incorrect notation such as 2+, 2+
or II, once only. 2
(ii) nitric acid/HNO3 / NO3–/nitrate;
Allow nitrogen from nitric acid/nitrate but not just nitrogen. 1
(b) (i) 0.100 × 0.0285; 2.85 × 10–3 (mol);
(ii) 2.85 × 10–3 (mol); 1
(iii) (63.55 × 2.85 × 10–3) = 0.181 g;
Allow 63.5. 1
(iv)
= ×100 456 . 0
181 . 0
39.7 % 1
(v)
= × −
100 2
. 44
7 . 39 2 . 44
10/10.2 %;
Allow 11.3 % i.e. percentage obtained in (iv) is used to
divide instead of 44.2 %. 1
(c) Brass has:
delocalized electrons / sea of mobile electrons / sea of electrons free to move;
No mark for just “mobile electrons”. 1
18. B
19. B
20. A
21. D
22. (a) n(HCl)(= 0.200 mol dm–3 × 0.02720 dm3) = 0.00544 / 5.44 × 10–3 (mol); 1
(b) n(HCl) excess (= 0.100 mol dm–3 × 0.02380 dm3) = 0.00238 / 2.38 × 10–3 (mol);
Penalize not dividing by 1000 once only in (a) and (b). 1
(c) n(HCl) reacted (= 0.00544 – 0.00238) = 0.00306 / 3.06 × 10–3 (mol); 1
(d) 2HCl(aq) + CaCO3(s) → CaCl2(aq) + H2O(l) + CO2(g) / 2H+(aq) + CaCO3(s) → Ca2+(aq) + H2O(l) + CO2(g);
Award [1] for correct reactants and products. Award [1] if this equation correctly balanced. Award [1 max] for the following equations: 2HCl(aq) + CaCO3(s) → CaCl2(aq) + H2CO3(aq) 2H+(aq) + CaCO3(s) → Ca2+(aq) + H2CO3(aq)
Ignore state symbols. 2
(e) n(CaCO3) = (2
1n(HCl)) = 2
1 × 0.00306;
= 0.00153 / 1.53 × 10–3 (mol);
Award [2] for correct final answer. 2
M = 100.09 /100.1 (g mol–1);
Accept 100.
m(CaCO3)(= nM) = 0.00153 (mol) × 100.09 (g mol–1) = 0.153 (g);
%CaCO3
×
= 100
188 . 0
153 . 0
= 81.4 % / 81.5 %
Accept answers in the range 79.8 to 81.5 %.
Award [3] for correct final answer. 3
(g) only CaCO3 reacts with acid / impurities are inert/non-basic / impurities do not react with the acid / nothing else in the eggshell reacts with acid / no other carbonates;
Do not accept “all calcium carbonate reacts with acid”. 1
23. C 24. D 25. D 26. B 27. C 28. C
29. (a) nC = 01 . 12
7 . 81
= 6.80 and nH = 01 . 1
3 . 18
= 18.1;
ratio of 1: 2.67 /1: 2.7; C3H8;
No penalty for using 12 and 1. 3
(b) C3H8; 1
(c) (i) Br2 /bromine; UV/ultraviolet light;
Accept hf/hv/sunlight. 2
(ii) Cr2O72– /MnO4– and acidified/ H+ /H3O+;
Accept names.
heat / reflux; 2
(d) Initiation:
Br2→ 2Br•;
Propagation:
Br• + RCH3→ HBr + RCH2•; RCH2• + Br2→ RCH2Br + Br•; Termination: [1 max]
Br• + Br• → Br2;
RCH2• + Br• → RCH2Br; RCH2• + RCH2• → RCH2CH2R;
Award [1] for any termination step.
Accept radical with or without • throughout.
(e) (i) substitution and nucleophilic and bimolecular/two species in rate-determining step;
Allow second order in place of bimolecular. 1
(ii)
curly arrow going from lone pair/negative charge on O in OH– to C;
Do not allow curly arrow originating on H in OH–
curly arrow showing Br leaving;
Accept curly arrow either going from bond between C and Br to Br in bromoethane or in the transition state.
representation of transition state showing negative charge, square brackets and partial bonds;
Do not penalize if HO and Br are not at 180° to each other. Do not award M3 if OH ---- C bond is represented unless already penalized in M1.
Do not penalize the use of an incorrect alkyl chain in the mechanism. 3 30. C
31. D 32. A 33. C 34. C 35. B 36. B
37. (a) ester; 1
(b) amount of oil = 6 . 885
0 . 1013
= 1.144 mol;
amount of methanol = 05 . 32
0 . 200
= 6.240 mol;
since three mol of methanol react with one mol of vegetable oil the
amount of excess methanol = 6.204 – (3 × 1.144) = 2.808 mol; 3
(c) (i) rate of the forward reaction is equal to the rate of the reverse reaction / forward and reverse reactions occur and the concentrations of the
reactants and products do not change / OWTTE; 1
(ii) Kc = 3
3
[methanol] oil]
vegetable [
] [biodiesel glycerol]
[
× ×
(iii) to move the position of equilibrium to the right/product side /
increase the yield of biodiesel; 1
(iv) no effect (on position of equilibrium);
increases the rate of the forward and the reverse reactions equally (so equilibrium reached quicker) / it lowers Ea for both the forward and reverse reactions by the same amount / OWTTE;
No ECF for explanation. 2
(d) vegetable oil is mainly non-polar and methanol is polar / OWTTE; stirring brings them into more contact with each other / increase the frequency of collisions / OWTTE;
Do not allow simply mixing. 2
(e) (relative molecular mass of biodiesel, C9H36O2 = 296.55) maximum yield of biodiesel = 3.432 mol / 1018 g;
percentage yield = 1018
0 . 811
× 100 = 79.67 %;
Allow 80 % for percentage yield. 2
(f) the carbon dioxide was absorbed by plants initially so there is no net increase / vegetable oil is not a fossil fuel / vegetable oil is formed
from (atmospheric) carbon dioxide / OWTTE; 1
38. (a) (i) Fe2+→ Fe3+ + e–; 1
(ii) MnO4– + 8H+ + 5e–→ Mn2+ + 4H2O; 1
(iii) MnO4– + 5Fe2+ + 8H+→ Mn2+ + 5Fe3+ 4H2O;
Accept e instead of e–. 1
(b) (i) amount of MnO4– = 1000
6 . 11
× 0.0200 = 2.32 × 10–4 mol; 1
(ii) amount of Fe2+ = 5 × 2.32 × 10–4 = 1.16 × 10–3 mol; 1 (iii) mass of Fe2+ = 55.85 × 1.16 × 10–3 = 6.48 × 10–2 g;
percentage of Fe2+ in tablet =
43 . 1
10 48 .
6 × −2
× 100 = 4.53 %; 2
39. D 40. C 41. B 42. D
43. (a) carboxylic acid / carboxyl; ester;
aryl group / benzene ring / phenyl; 3
(b) (i) Mr(C7H6O3) = 138.13;
n =
= 13 . 138
15 . 3
2.28 × 10–2 (mol);
Award [2] for the correct final answer. 2
(ii) Mr(C9H8O4) = 180.17;
m = (180.17 × 2.28 × 10–2 =) 4.11(g);
Accept range 4.10–4.14
Award [2] for the correct final answer. 2
(iii) (percentage yield = 11 . 4
50 . 2
× 100 =) 60.8 %;
Accept 60–61 %. 1
(iv) 3;
(percentage uncertainty = 50 . 2
02 . 0
× 100 =) 0.80 %;
Allow 0.8 % 2
(v) sample contaminated with ethanoic acid / aspirin not dry / impure sample;
Accept specific example of a systematic error.
Do not accept error in reading balance/weighing scale.
Do not accept yield greater than 100 %. 1
44. C N O H
20.2 11.4 65.9 2.50
12.01 14.01 16.00 1.01
= 1.68 = 0.814 = 4.12 = 2.48 ;
814 . 0
68 . 1
= 2 814 . 0
814 . 0
= 1 814 . 0
12 . 4
= 5 814 . 0
48 . 2
= 3 ;
C2NO5H3;
No penalty for use of 12, 1 and/or 14.
Award [1 max] if the empirical formula is correct, but no working shown. 3 45. C
46. B 47. D 48. D 49. C
50. (a) MnO4–(aq) + 5Fe2+(aq) + 8H+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
Award [2] if correctly balanced.
Award [1] for correctly placing H+ and H2O.
Award [1 max] for correct balanced equation but with electrons shown.
Ignore state symbols. 2
(b) Fe2+/ iron(II);
Do not accept iron. 1
(c) n = 2.152 × 10–2 × 2.250 × 10–2; 4.842 × 10–4 (mol);
Award [1] for correct volume
Award [1] for correct calculation. 2
(d) 1 mol of MnO4– reacts with 5 mol of Fe2+; 5 × 4.842 × 10–4 = 2.421 × 10–3 (mol); (same number of moles of Fe in the iron ore)
Allow ECF from part (a) and (c) provided some mention of mole ratio
is stated. 2
(e) 2.421 × 10–3 × 55.85 = 0.1352(g);
3682 . 0
1352 . 0
× 100 = 36.72 %;
Allow ECF from part (d). 2
51. C 52. D
53. B 54. C 55. C 56. C
57. (a) 0.600 mol Al(OH)3≡ (1.5)(0.600) mol H2SO4/0.900 mol H2SO4 needed, but only 0.600 mol used;
H2SO4 limiting reactant; 2
Some working must be shown in order to score the second point.
(b) 0.200 mol Al2(SO4)3;
68.4(g); 2
Penalize incorrect units.
(c) 0.200 mol; 1
Use ECF from (a).
(d) A Brønsted-Lowry acid is a proton/H+ donor;
A Lewis base is an electron-pair donor; 2
(e) H2CO3 and carbonic acid / CH3COOH and ethanoic acid; 1
Accept any other weak acid and correct formula.
58. empirical formula = CN;
Working must be shown to get point.
Mr = 51.9 (g mol–1);
