06 The Mole Concept and Stoichiometry, Set A Answer Key To print: Ctrl + P, paper A4, margins Default.
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06

The Mole Concept and Stoichiometry

Set AAnswer key
A

From the lesson

23 marks
1

Let the abundance of 79Br be x%.

79.90 = 79x + 81(100 − x)100[1]

7990 = 8100 − 2x

x = 55.0, so 55.0% 79Br (and 45.0% 81Br)[1]

2
CH
mass in 100 g / g85.714.3
amount / mol85.7 ÷ 12.0 = 7.14214.3 ÷ 1.0 = 14.30
ratio12

Empirical formula CH2 (Mr = 14.0)[1]

n = 56.014.0 = 4, so the molecular formula is C4H8[1]

3

massC = 12.044.0 × 0.880 = 0.2400 g, massH = 2.018.0 × 0.360 = 0.04000 g[1]

massO = 0.600 − 0.2400 − 0.04000 = 0.3200 g[1]

nC : nH : nO = 0.02000 : 0.04000 : 0.02000 = 1 : 2 : 1, so the empirical formula is CH2O[1]

4

nCaCO3 = 5.00100.1 = 0.04995 mol, nHCl = 50.01000 × 1.00 = 0.05000 mol[1]

0.04995 mol of CaCO3 needs 0.09990 mol of HCl, but only 0.05000 mol is present, so HCl is limiting.[1]

nCO2 = ½ × 0.05000 = 0.02500 mol

VCO2 = 0.02500 × 24.0 = 0.600 dm3[1]

5

CxHy + (x + y4)O2 → xCO2 + y2H2O

VCO2 = 110 − 50 = 60 cm3 = 20x, so x = 3[1]

VO2 used = 150 − 50 = 100 cm3 = 20(x + y4), so y = 8[1]

The hydrocarbon is C3H8 (propane).[1]

NoteThe gas left after NaOH(aq) is the unreacted O2. Water is a liquid at room temperature and pressure, so it takes no part in the gas volume.

6

nHCl added = 0.05000 mol, nHCl left over = nNaOH = 20.01000 × 0.500 = 0.01000 mol[1]

nHCl reacted = 0.04000 mol, so nCaCO3 = ½ × 0.04000 = 0.02000 mol[1]

massCaCO3 = 0.02000 × 100.1 = 2.002 g

% purity = 2.0022.50 × 100% = 80.1%[1]

7
(a)
2x + 7(−2) = −2, so x = +6[1]
(b)
2x + 3(−2) = −2, so x = +2[1]
(c)
x + 4(+1) = +1, so x = −3[1]
8

Reduction: Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O

Oxidation (× 6): Fe2+ → Fe3+ + e−[1]

Overall: Cr2O72− + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+[1]

Exam tipCheck that both atoms and total charge balance: left −2 + 14 + 12 = +24, right 6 + 18 = +24.

9

nSO32− = 25.01000 × 0.100 = 2.500 × 10−3 mol. Each SO32− loses 2 electrons as the oxidation number of S increases from +4 to +6, so 5.000 × 10−3 mol of electrons are lost.[1]

noxidant = 20.01000 × 0.0500 = 1.000 × 10−3 mol, so each M gains 5 electrons. Final oxidation number of M = +5 − 5 = 0[1]

B

Practice

15 marks
10

CRelative atomic mass of oxygen = average mass of one atom of oxygen ÷ (1/12 × mass of one atom of carbon-12)[1]

ABy definition, one mole of a substance contains exactly 6.02 × 1023 elementary entities. A is incorrect because of the phrase "same number of atoms as there are in 12.000 g of carbon-12". BRelative isotopic mass is the mass of one mole of atoms of an isotope relative to 1/12 the mass of one mole of 12C atoms. B is incorrect because the numerator used is "average mass of all isotopes of lithium". CRelative atomic mass is the average mass of one mole of atoms of an element relative to 1/12 the mass of one mole of 12C atoms or the average mass of one atom of an element relative to 1/12 the mass of one 12C atom. C is correct. DRelative molecular mass is the average mass of one mole of molecules relative to 1/12 the mass of one mole of 12C atoms. D is incorrect because the numerator used is "average mass of one atom of E".
11

B37.5 cm3[1]

Dilution does not change the amount: ciVi = cfVf

(0.280)(50) = (0.160)Vf, so Vf = 87.5 cm3

Volume of water added = 87.5 − 50.0 = 37.5 cm3

12
(a)

Reduction: PbO2 + 2e− → PbO22−

Oxidation: Cr(OH)3 + 5OH− → CrO42− + 4H2O + 3e−[1]

Overall: 2Cr(OH)3 + 10OH− + 3PbO2 → 2CrO42− + 8H2O + 3PbO22−[1]

(b)

Volumes used: 25.55, 25.40, 25.30 cm3 (2 d.p.)[1]

Average volume of Fe2+ used = ½(25.40 + 25.30) = 25.35 cm3 (the two concordant titres)

Amount of Fe2+ reacted = 25.351000 × 0.150 = 3.80 × 10−3 mol[1]

(c)
% uncertainty = 0.05 + 0.0525.35 × 100% = 0.394%[1]
(d)

6Fe2+ ≡ Cr2O72− ≡ 2CrO42−

Amount of Cr2O72− in 25.0 cm3 = 3.803 × 10−36 = 6.338 × 10−4 mol

Amount of CrO42− in 250 cm3 = 6.338 × 10−4 × 2 × 25025.0 = 0.01268 mol[1]

2CrO42− ≡ 3PbO2, so amount of PbO2 = 0.01268 × 32 = 0.01901 mol

Mass of PbO2 = 0.01901 × (207.2 + 32.0) = 4.548 g

% by mass of PbO2 = 4.5485.00 × 100% = 91.0%[1]

13
(a)

Oxidation: Mn2+ + 3OH− → Mn(OH)3 + e−[1]

Overall: 4Mn2+ + 8OH− + 2H2O + O2 → 4Mn(OH)3[1]

(b)

(i) 1O2 ≡ 4Mn(OH)3 ≡ 2I2 ≡ 4Na2S2O3[1]

(ii) Amount of Na2S2O3 = 25.001000 × 0.00100 = 2.50 × 10−5 mol

Amount of dissolved O2 = 2.50 × 10−54 = 6.25 × 10−6 mol[1]

(iii) Mass of O2 = 6.25 × 10−6 × 32.0 = 2.00 × 10−4 g = 0.200 mg

DOC = 0.200 × 100025.0 = 8.00 mg dm−3[1]

8.00 mg dm−3 is more than 5 mg dm−3, so fish can survive in the river.[1]