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Atoms, Molecules and Stoichiometry_2014 (Suggested Solutions) 1 Topic: Atoms, Molecules and Stoichiometry (Suggested Solutions) Section A: Multiple Choice Questions 1. D CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) C 2 H 6 (g) + 7/2O 2 (g) → 2CO 2 (g) + 3 H 2 O(l) The gas absorbed by KOH is CO 2 . CO 2 (g) + 2OH - (aq) → CO 3 2- (aq) + H 2 O(l) Volume of CO 2 produced by CH 4 = 10 cm 3 Volume of CO 2 produced by C 2 H 6 = 20 cm 3 Total volume = 10 + 20 = 30 cm 3 2. D Molar ration of M:O = 0.02 : (2x0.025) = 2 : 5 3. C 4KClO 3 → 3KClO 4 + KCl Amount of KClO 4 produced = ¾ x 0.1 = 0.0750 mol 4. B C x H y + (x+y/4)O 2 xCO 2 + y/2H 2 O 1 volume of C x H y produces x volumes of CO 2 and y/2 volumes of H 2 O vapour. Therefore, from the data, x=3 and y/2 = 2. 5. D Aqueous NaOH absorbs CO 2 and the remaining gas is N 2 . Amount of urea = 0.120/60 = 2.00 x 10 -3 mol Volume of N 2 = 2 x (2.00 x 10 -3 ) x 24000 = 96.0 cm 3 6. B 1 mol of H - gives 1 mol of H 2 . The salt with the greatest amount of H - will therefore give the highest amount of H 2 . 7. D 2H 2 + O 2 → 2H 2 O O 2 is the limiting reagent. Therefore, max decrease in volume = 10 + 20 = 30 cm 3 8. D Let x be the number of C atoms per molecule of X. Therefore, 1 mole of X will produce x moles of CO 2 when completely burnt in O 2 . Amount of X used = 0.112/22.4 = 5.00 x 10 -3 mol Amount of CO 2 produced = 0.88/ (12.0 + 2(16.0)) = 2.00 x 10 -2 mol (5.00 x10 -4 )x = 2.00 x 10 -2 x = 4 9. A Amount of H + = amount of OH - = 0.1 x 24/1000 = 0.0024 mol (in 20 cm 3 ) Amount of Cl - = 0.1 x 48/1000 = 0.0048 mol (in 20 cm 3 ) Ratio of H + : Cl - = 0.0024 : 0.0048 = 1 : 2

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  • Atoms, Molecules and Stoichiometry_2014 (Suggested Solutions)

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    Topic: Atoms, Molecules and Stoichiometry (Suggested Solutions) Section A: Multiple Choice Questions 1. D CH4 (g) + 2O2(g) CO2(g) + 2H2O(l) C2H6(g) + 7/2O2(g) 2CO2(g) + 3 H2O(l) The gas absorbed by KOH is CO2. CO2(g) + 2OH

    -(aq) CO32-(aq) + H2O(l)

    Volume of CO2 produced by CH4 = 10 cm3

    Volume of CO2 produced by C2H6 = 20 cm3

    Total volume = 10 + 20 = 30 cm3 2. D Molar ration of M:O = 0.02 : (2x0.025) = 2 : 5 3. C 4KClO3 3KClO4 + KCl Amount of KClO4 produced = x 0.1 = 0.0750 mol 4. B CxHy + (x+y/4)O2 xCO2 + y/2H2O 1 volume of CxHy produces x volumes of CO2 and y/2 volumes of H2O vapour. Therefore, from the data, x=3 and y/2 = 2. 5. D Aqueous NaOH absorbs CO2 and the remaining gas is N2. Amount of urea = 0.120/60 = 2.00 x 10-3 mol Volume of N2 = 2 x (2.00 x 10

    -3) x 24000 = 96.0 cm3 6. B 1 mol of H- gives 1 mol of H2. The salt with the greatest amount of H

    - will therefore give the highest amount of H2. 7. D 2H2 + O2 2H2O O2 is the limiting reagent. Therefore, max decrease in volume = 10 + 20 = 30 cm

    3 8. D Let x be the number of C atoms per molecule of X. Therefore, 1 mole of X will produce x moles of CO2 when completely burnt in O2. Amount of X used = 0.112/22.4 = 5.00 x 10-3 mol Amount of CO2 produced = 0.88/ (12.0 + 2(16.0)) = 2.00 x 10

    -2 mol (5.00 x10-4)x = 2.00 x 10-2 x = 4 9. A Amount of H+ = amount of OH- = 0.1 x 24/1000 = 0.0024 mol (in 20 cm3) Amount of Cl- = 0.1 x 48/1000 = 0.0048 mol (in 20 cm3) Ratio of H+ : Cl- = 0.0024 : 0.0048 = 1 : 2

  • Atoms, Molecules and Stoichiometry_2014 (Suggested Solutions)

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    10. A H+ + OH- H2O Amount of H+ = amount of OH- = conc x volume = 1.0 x 10-2 x 25/1000 =2.50 x 10-4 mol 2 H+ are required to remove 1 Ca2+ Amount of Ca2+ = amount of H+ = 1.25 x 10-4 mol [Ca2+] = 1.25 x 10-4/ (50/1000) = 2.50 x 10-3 mol dm-3 11. A CH4 + 2O2 CO2 + 2H2O CH4 + 3/2O2 CO + 2H2O Volume of O2 required for first equation = (y x 0.99) x 2 dm

    3 Volume of O2 required for second equation = (y x 0.01) x 3/2 dm

    3 Hence total volume of O2 required = [(y x 0.99) x 2] + [(y x 0.01) x 3/2] = 2y 0.01y/2 dm3

    13. D 2AgNO3 + BaCl2 2AgCl + Ba(NO3)2 Amount of BaCl2 = conc x vol = 0.20 x 20/1000 = 4 x 10

    -3 mol Amount of AgNO3 = 2 x 4 x 10

    -3 Volume of AgNO3 = 8 x 10

    -3/ 0.10 = 0.08 dm3 = 80 cm3

    14. C

    mass of CaBr2 in 100 tonnes of solution = (52/100 x 100) tonnes

    Mr of CaBr2 = 40 + 2(80) = 200

    mass of bromine = 100100

    52

    200

    )80(2 = 52

    200

    160 tonnes

    15. B

    In B: H2 2H+ + 2e

    -

    Amount of e- = 2 x amt of H2 = 2 x = 1 mol

    16. D

    mole ratio Hg : lipoyl group = 1:1

    amount of Hg = amount of lipoyl group = (5.0 x 1.0 x 10-8) mol = 5.0 x 10-8 mass of Hg = n Ar = (5.0 x 10

    -8 x 200) g = 1.0 x 10

    -5 g

    17. C

    CxHy + (x+y/4)O2 xCO2 + y/2H2O 10 cm

    3 (70-20) cm

    3 30 cm

    3

    1 vol 5 vol 3 vol

    x = 3; x + y/4 = 5 y = 8 18. D CxHy + (x+y/4)O2 xCO2 + y/2 H2O Amount of CO2 = 0.352/44.0 = 0.00800 mol Amount of H2O = 0.072/18.0 = 0.00400 mol y=x E.F. CxHx

  • Atoms, Molecules and Stoichiometry_2014 (Suggested Solutions)

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    19. C Taking X = OH, the reaction becomes:

    Zn(C2H5)2 + 2H-OH Zn(OH)2 + 2C2H6

    20. A It contains the same number of atoms as 1 mol of hydrogen atoms. One mole of any substance contains the same number (Avocadros constant 6.02 x 1023) of particles (atoms, molecules, ions). 21. C

    CS2(g) + 3O2(g) CO2(g) + 2SO2(g) 20 cm3 60 cm3 20 cm3 40 cm3 Volume of unreacted O2 = 100 60 = 40 cm

    3 Final volume of gas = (40 +20 +40) = 100 cm3 Both CO2 and SO2 (acidic gases) are absorbed by aq alkali.

    % dissolved in alkali = 100100

    4020

    = 60 %

    22. A Relative atomic mass is the ratio of the mass of one atom of the element to 1/12 of the mass of an atom of 12C isotope, expressed on the 12C scale. 23. D

    CH2O + O2 CO2 + H2O Amount of CO2 = amount of CH2O

    = 0.30

    100080.1

    0.16)0.1(20.12

    100080.1

    = 60.0

    mass of CO2 = n Mr = [60 x (12 + (2x16))] = (60.0 x 44) = 2640 g = 2.64 kg 24. D CnH2n+2 + ((3n+1)/2)O2 nCO2 + (n+1)H2O Residual gas, V = CO2 and unreacted O2 at r.t.p. For 10 cm3 of CH4 CO2 = 10 cm3 O2 unreacted = 70-20 = 50 cm

    3 V = 60 cm3 Only option D starts with 60 cm3. 25. B

    CxHyNz + O2 xCO2 + y/2H2O + z/2N2 10 cm3 of CxHyNz, when burnt in O2, gives 10x cm

    3 of CO2 and 10z/2 = 5z cm3

    of N2 Hence, x = 2 and z = 1

  • Atoms, Molecules and Stoichiometry_2014 (Suggested Solutions)

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    Section B: Structured Questions 26.

    Barium oxide: BaO + H2SO4 BaSO4 + H2O

    Barium peroxide: BaO2 + H2SO4 BaSO4 + H2O2 (a) Barium sulfate Amount = 1.63/ 233.1 = 0.00699 mol

    (b) 2MnO4- 5H2O2

    Amount of MnO4- = (24.5/1000) x 0.02 = 0.000490 mol

    Amount of H2O2 = 0.00049 x 2.5 = 0.00123 mol (c) Amount of H2O2 = amt of BaO2 = 0.00123 mol Mass of BaSO4 = 0.001225 x 169.0 = 0.207 g Amount of BaSO4 fr BaO = 0.00699 0.001225 = 0.00577 mol Mass of BaO = 0.00577 x 153.0 = 0.883 g 27. (a) Na2CO3 (s) + 2 HCl (aq) 2 NaCl (aq) + CO2(g) + H2O (l) (b) Amount of HCl in 50 cm3 = 0.05 x 0.1 = 0.00500 mol Amount of pure Na2CO3 = 0.005 x = 0.00250 mol Mass of pure Na2CO3 = 0.0025 x 106 = 0.265 g Percentage purity of Na2CO3 = (0.265 / 1.20) x 100 = 22.1%. 28. (a) Molar mass of compound A = 17.6 0.200 = 88.0 g mol-1

    (b)

    Element C H N

    % mass 54.5 13.6 31.8

    RAM 12.0 1.0 14.0

    % mass/ RAM 4.54 13.6 2.27

    Ratio 2 6 1

    Empirical formula is C2H6N. (c) Let the molecular formula be (C2H6N)n. n = 88.0 (2 12.0 + 6 1.0 + 14.0) = 2 (whole number) Molecular formula is C4H12N2 29.

    (a) 8H+ + MnO4- + 5Fe2+ Mn2+ + 5 Fe3+ + 4H2O

    (b)

    mol 10 x 3.58

    0.0250 x 1000

    14.30MnO ofAmount

    4-

    4

    2 3Amount of Fe (in 1 dm of solution)

    -4 10005 x 3.575 x 10 x 25

    0.0715 mol

    (c) g 10.9

    16.0) 4 32.1 (55.8 x 0.715solution) of dm 1 (inFeSO of Mass 34

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    30. (a) Ca(OH)2 + 2HCl CaCl2 + 2H2O (b) Amount of HCl used = 26.50 10-3 0.100 = 2.65 10-3 mol (c) Amount of Ca(OH)2 in 25.0 cm

    3 = 2.65 10-3 1/2 = 1.325 10-3 = 1.33 10-3 mol (d) Concentration of Ca(OH)2 = (1.33 10

    -3) (25.0 10-3) = 0.0532 mol dm-3 (e) Percentage purity of sample = [(0.0532 200 10-3 74.1) 3.0] 100 % =26.3% 31. (a) M1V1 = M2V2 1.0 x 25 = M x 250 M = 0.100 mol dm-3 Or Amount of FeSO4 in 25.0 cm

    3 = 25/1000 x 1.00 = 0.0250 mol Conc of diluted FeSO4 = 0.025 / (250/1000) = 0.100 moldm

    -3

    (b) 1000

    10x 0.100 = 0.00100 mol

    (c) 1000

    10 x 0.0250 = 0.000250 mol

    (d) MnO4- : Fe2+

    0.000250 : 0.00100 1 : 4 Fe2+ oxidizes to Fe3+ by losing 1 mol of e (ie Fe2+ Fe3+ + e). Hence, for 4 mols of Fe2+, 4 mols of electron will be gained by 1 mol of MnO4

    -. So, final oxidation no. of Mn = +7 - 4 = +3

    32. (a) moles (S2O32-) = 0100.04.0

    1000

    00.25

    moles (I2) = 31000.5

    2

    0100.0

    (b) moles (Cr2O72-) =

    33

    1067.13

    1000.5

    (c) moles (Na3CrO4) = moles (CrO43-) = 3 x 1.67 x 10-3 = 5.00 x 10-3

    (d) mass of Na3CrO4 = (5.00 x 10

    -3) (185) = 0.925g

    % by mass of Na3CrO4 = %0.50%10085.1

    925.0

    33. a)

    C H O

    m/Ar 40.91/12.0 4.55/1 54.54/16

    ratio 3.409 4.55 3.409

    simplest ratio 1.00 = 3 1.34 = 4 1.00 = 3

    EF = C3H4O3 b) Let MF = (C3H4O3 )n

    88n = 176, hence n = 2; MF = C6H8O6

  • Atoms, Molecules and Stoichiometry_2014 (Suggested Solutions)

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    c) Mass of vit C = 19.1/100 x 1.28 = 0.244 g Amount of vit.C = 0.244/176 = 1.39 x 10-3 mol Conc of vit C = 1.39 x 10-3/0.25 = 5.56 x 10-3 mol dm-3

    34. C4Hy(g) + (4+ y/4)O2(g) 4CO2(g) + y/2 H2O(g) Change in volume -10 -(4+ y/4)10 40 (y/2)10 =5y 5y + 40 -10-(4+ y/4)10 = 10 y = 8

    35. [Mx . 25] /[My.20] = 5/1 Mx/My = 4 [Mx.25]/[My.V] = 3/1 V = [25 x 4]/3 = 33.3 Ans = 33.3 cm3

    36.

    (a) 5SO32- + 2 MnO4

    - + 6H+ 2Mn2+ + 3H2O + 5SO42-

    (b) Amount of MnO4- = 0.02 x 0.03 = 6.00 x 10-4 mol

    Amount of Na2SO3 = 5/2 x 6 x 10-4= 1.50 x 10-3 mol

    Mass of in 1 kg of meat = 1.5 x 10-3 x 126 = 1.89 x 10-1g (c) Mass of in 106 g meat = 1.89 x 10-1 x 103 g = 189 g i.e. 189 p.p.m. (d) 1 mol SO3

    2- 1 mol SO2 2 mol HCl Amount of HCl = 2 x 1.5 x 10-3 = 3.00 x 10-3 mol Vol. of HCl needed = (3 x 10-3) 0.1 = 3.00 x 10-2 dm3 or 30.0 cm3

    37.

    (i) Amount of H2 gas = 23

    4190 = 182.2 = 182 mol

    Mass of H2 = 182.2 x 2 = 364 g (ii) Volume of air displaced = 4190 cm3

    there should be 182 moles of air. Mass of air displaced = 182.2 x 29 = 5283.8 g = 5280 g (iii) Upthrust = mass of air displaced = mass of H2 + load Load = 5283.8 364.4 = 4919.4 g = 4.92 kg

    38. (a) (i) (CH2O)n + nO2(g) nCO2(g) + nH2O(l) (ii) Mr of carbohydrate = 30 n

    Amount of carbohydrate = n30

    1200 =

    n

    40 mol

    Amount of O2 = n

    40 x n = 40.0 mol

    (iii) Assuming that there are 365 days per year. Amount of O2 required = 40 x (70x365) = 1.02 x 10

    6 mol

    (b) (i) C8H18(l) + 2

    25O2(g) 8CO2(g) + 9H2O(l)

    (ii) Mass of octane = density x volume = 0.7 x (6 x 103) = 4200g

    Amount of octane = Mr

    mass =

    114

    4200= 36.8 mol

    (iii) Amount of O2 required = 2

    25 x 36.84 = 461 mol

  • Atoms, Molecules and Stoichiometry_2014 (Suggested Solutions)

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    (c) To run 100 km, 460.5 mol of O2 are required

    using 1.022 x 106 mol of O2,

    Distance travelled = 5.460

    10022.1 6 x 100 = 2.22 x 105 km

    39. (a) (i) CaCO3 + SO2 CaSO4 + CO2

    (ii) CaSO3 + 2

    1O2 CaSO4

    (b) (i) CaCO3 + SO2 CaSO3 + CO2

    Amount of SO2 = amount of CaCO3 =

    1.100

    10)103 65

    mass of SO2 = (

    1.100

    10)103 65 x 64.1) = 1.92 x 1011 g = 1.92 x 105 tonnes

    (ii) CaSO3 + 2

    1O2 CaSO4

    CaCO3 CaSO3 CaSO4 Amount of CaSO4 = amount of CaCO3

    mass of CaSO4 = (

    1.100

    10)103 65 x 136.2) = 4.08 x 1011 g = 4.08 x 105 tonnes

    (c) 90 % of SO2 removed from waste gases;

    10 % of SO2 is released into atmosphere

    mass of SO2 = (51092.1

    100

    10 ) 1.92 x 104 g