A Data Booklet may be used. Molar volume of a gas = 24.0 dm3 mol−1 at r.t.p.
Source: Learners' Alley, The Mole Concept slides
Source: Learners' Alley, The Mole Concept slides
Source: Learners' Alley, The Mole Concept slides
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
Identify the limiting reagent, and calculate the volume of CO2 formed at r.t.p.Source: Learners' Alley, The Mole Concept slides
Source: Learners' Alley, The Mole Concept slides
Source: Learners' Alley, The Mole Concept slides
Determine the oxidation number of
Source: Learners' Alley, The Mole Concept slides
Source: Learners' Alley, The Mole Concept slides
Source: Learners' Alley, The Mole Concept slides
CommonWhich statement is correct?
Source: 2024 RVHS J2 Prelim P1 Q2 (also 2019 RI J1 Promo P1 Q1)
What volume of distilled water must be added to 50 cm3 of 0.280 mol dm−3 KMnO4 to dilute it to 0.160 mol dm−3?
Source: IJC 2017 Promo, via EJC 2022 Mole Concept Topical Revision, Q10
CommonLead(IV) oxide, PbO2, is a strong oxidising agent. Under basic conditions, solid PbO2 reacts with solid Cr(OH)3 to form CrO42− and PbO22− ions.
A 5.00 g sample containing PbO2 was reacted with excess Cr(OH)3 in alkaline solution, and the resulting solution of CrO42− was made up to 250 cm3. A 25.0 cm3 aliquot was acidified, converting the CrO42− to Cr2O72−, and titrated against 0.150 mol dm−3 Fe2+.
2CrO42− + 2H+ → Cr2O72− + H2O 6Fe2+ + Cr2O72− + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O
| titration number | 1 | 2 | 3 |
|---|---|---|---|
| final burette reading / cm3 | 25.55 | 33.75 | 40.90 |
| initial burette reading / cm3 | 0.00 | 8.35 | 15.60 |
| volume of Fe2+(aq) used / cm3 |
Source: MJC 2017 Promo, via EJC 2022 Mole Concept Topical Revision, Q4
ChallengeIf the dissolved oxygen concentration (DOC) in a river falls below 5 mg dm−3, most fish cannot survive. In the Winkler method, river water is shaken with excess alkaline Mn2+, which oxygen oxidises to a pale brown precipitate of Mn(OH)3.
The Mn(OH)3 is then reacted with excess KI, and the iodine formed is titrated with sodium thiosulfate.
2KI + 2Mn(OH)3 → I2 + 2Mn(OH)2 + 2KOH 2Na2S2O3 + I2 → Na2S4O6 + 2NaI
Source: CJC 2018 Promo, via EJC 2022 Mole Concept Topical Revision, Q5
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]
| C | H | |
|---|---|---|
| mass in 100 g / g | 85.7 | 14.3 |
| amount / mol | 85.7 ÷ 12.0 = 7.142 | 14.3 ÷ 1.0 = 14.30 |
| ratio | 1 | 2 |
Empirical formula CH2 (Mr = 14.0)[1]
n = 56.014.0 = 4, so the molecular formula is C4H8[1]
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]
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]
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.
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]
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.
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]
CRelative atomic mass of oxygen = average mass of one atom of oxygen ÷ (1/12 × mass of one atom of carbon-12)[1]
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
Reduction: PbO2 + 2e− → PbO22−
Oxidation: Cr(OH)3 + 5OH− → CrO42− + 4H2O + 3e−[1]
Overall: 2Cr(OH)3 + 10OH− + 3PbO2 → 2CrO42− + 8H2O + 3PbO22−[1]
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]
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]
Oxidation: Mn2+ + 3OH− → Mn(OH)3 + e−[1]
Overall: 4Mn2+ + 8OH− + 2H2O + O2 → 4Mn(OH)3[1]
(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]