The Mole Concept, Avogadro, Concentration, Gas Volumes and Stoichiometry

1. What amount of substance is present in 9.80 g of sulfuric acid, H2SO4?

[Ar: H, 1; S, 32; O, 16]

A 0.0100 mol
B 0.100 mol
C 0.980 mol
D 9.80 mol

Answer: B

Mr of H2SO4 = 2(1) + 32 + 4(16) = 98
Moles = mass / Mr = 9.80 / 98 = 0.100 mol

A wrong: Divides by 980 instead of 98.
B right: 0.100 mol.
C wrong: Too high.
D wrong: Confuses mass with moles.

2. How many oxygen atoms are present in 0.250 mol of carbon dioxide molecules?

[Avogadro constant = 6.02 x 10^23 mol–1]

A 1.51 x 10^23
B 3.01 x 10^23
C 6.02 x 10^23
D 1.20 x 10^24

Answer: B

Each CO2 molecule contains 2 oxygen atoms.
0.250 mol CO2 contains 0.500 mol oxygen atoms.
Atoms = 0.500 x 6.02 x 10^23 = 3.01 x 10^23

A wrong: Counts only CO2 molecules, not oxygen atoms.
B right: 3.01 x 10^23.
C wrong: Counts 1 mol oxygen atoms.
D wrong: Too high.

3. What is the volume of 0.750 mol of nitrogen gas at r.t.p.?

[molar gas volume at r.t.p. = 24 dm3/mol]

A 0.0313 dm3
B 12.0 dm3
C 16.0 dm3
D 18.0 dm3

Answer: D

Volume = moles x 24
Volume = 0.750 x 24 = 18.0 dm3

A wrong: Divides wrongly.
B wrong: Too low.
C wrong: Incorrect multiplication.
D right: 18.0 dm3.

4. Calcium carbonate decomposes when heated.

CaCO3 → CaO + CO2

What mass of carbon dioxide is produced from 20.0 g of pure calcium carbonate?

[Ar: Ca, 40; C, 12; O, 16]

A 4.40 g
B 8.80 g
C 11.2 g
D 20.0 g

Answer: B

Mr CaCO3 = 100
Moles CaCO3 = 20.0 / 100 = 0.200 mol
CaCO3 : CO2 = 1 : 1
Moles CO2 = 0.200 mol
Mass CO2 = 0.200 x 44 = 8.80 g

A wrong: Half the correct answer.
B right: 8.80 g.
C wrong: Uses gas-volume thinking wrongly.
D wrong: Product mass is not equal to reactant mass.

5. What is the concentration of a solution containing 0.250 mol of sodium hydroxide in 500 cm3 of solution?

A 0.00050 mol/dm3
B 0.125 mol/dm3
C 0.500 mol/dm3
D 2.00 mol/dm3

Answer: C

500 cm3 = 0.500 dm3
Concentration = moles / volume
= 0.250 / 0.500
= 0.500 mol/dm3

A wrong: Bad cm3 to dm3 conversion.
B wrong: Multiplies instead of dividing.
C right: 0.500 mol/dm3.
D wrong: Divides the wrong way round.

6. What volume of 0.200 mol/dm3 hydrochloric acid contains 0.0500 mol of HCl?

A 25.0 cm3
B 100 cm3
C 250 cm3
D 400 cm3

Answer: C

Volume = moles / concentration
= 0.0500 / 0.200
= 0.250 dm3
= 250 cm3

A wrong: Too small.
B wrong: Contains only 0.0200 mol.
C right: 250 cm3.
D wrong: Too large.

7. 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide reacts exactly with hydrochloric acid.

NaOH + HCl → NaCl + H2O

How many moles of hydrochloric acid react?

A 0.000250 mol
B 0.00250 mol
C 0.0250 mol
D 2.50 mol

Answer: B

Moles NaOH = 0.100 x 25.0/1000
= 0.00250 mol

NaOH : HCl = 1 : 1
Moles HCl = 0.00250 mol

A wrong: Ten times too small.
B right: 0.00250 mol.
C wrong: Did not convert cm3 to dm3.
D wrong: Impossible overestimate.

8. 20.0 cm3 of 0.150 mol/dm3 sulfuric acid reacts with sodium hydroxide.

H2SO4 + 2NaOH → Na2SO4 + 2H2O

How many moles of sodium hydroxide are needed?

A 0.00150 mol
B 0.00300 mol
C 0.00600 mol
D 0.300 mol

Answer: C

Moles H2SO4 = 0.150 x 20.0/1000
= 0.00300 mol

H2SO4 : NaOH = 1 : 2
Moles NaOH = 0.00600 mol

A wrong: Too low.
B wrong: Gives acid moles only.
C right: 0.00600 mol.
D wrong: Does not convert volume properly.

9. What mass of magnesium oxide is formed when 2.40 g of magnesium reacts completely with oxygen?

2Mg + O2 → 2MgO

[Ar: Mg, 24; O, 16]

A 1.60 g
B 2.40 g
C 4.00 g
D 8.00 g

Answer: C

Moles Mg = 2.40 / 24 = 0.100 mol
Mg : MgO = 1 : 1
Moles MgO = 0.100 mol
Mr MgO = 40
Mass MgO = 0.100 x 40 = 4.00 g

A wrong: Uses oxygen mass only.
B wrong: Assumes product mass equals magnesium mass.
C right: 4.00 g.
D wrong: Doubles the correct answer.

10. What volume of carbon dioxide is produced at r.t.p. when 0.0500 mol of calcium carbonate reacts completely with excess hydrochloric acid?

CaCO3 + 2HCl → CaCl2 + H2O + CO2

[molar gas volume at r.t.p. = 24 dm3/mol]

A 0.600 dm3
B 1.20 dm3
C 2.40 dm3
D 12.0 dm3

Answer: B

CaCO3 : CO2 = 1 : 1
Moles CO2 = 0.0500 mol
Volume CO2 = 0.0500 x 24 = 1.20 dm3

A wrong: Uses 12 dm3/mol.
B right: 1.20 dm3.
C wrong: Doubles the correct answer.
D wrong: Too large.

Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total A Grades, 11 World Records and 7 Distinctions, Educate A Change.

11. 2.40 dm3 of hydrogen reacts with excess oxygen at r.t.p.

2H2 + O2 → 2H2O

What volume of oxygen is required at r.t.p.?

A 1.20 dm3
B 2.40 dm3
C 3.60 dm3
D 4.80 dm3

Answer: A

Hydrogen : oxygen = 2 : 1
2.40 dm3 H2 needs 1.20 dm3 O2.

A right: 1.20 dm3.
B wrong: Assumes 1 : 1.
C wrong: Adds volumes.
D wrong: Doubles hydrogen volume.

12. 50.0 cm3 of methane is completely burned in oxygen.

CH4 + 2O2 → CO2 + 2H2O

What volume of oxygen is required, measured at the same temperature and pressure?

A 25.0 cm3
B 50.0 cm3
C 100 cm3
D 200 cm3

Answer: C

Methane : oxygen = 1 : 2
50.0 cm3 CH4 needs 100 cm3 O2.

A wrong: Half the methane volume.
B wrong: Assumes 1 : 1.
C right: 100 cm3.
D wrong: Too high.

13. 30.0 cm3 of ethene is completely burned in oxygen.

C2H4 + 3O2 → 2CO2 + 2H2O

What volume of carbon dioxide is formed, measured at the same temperature and pressure?

A 15.0 cm3
B 30.0 cm3
C 60.0 cm3
D 90.0 cm3

Answer: C

Ethene : carbon dioxide = 1 : 2
30.0 cm3 C2H4 forms 60.0 cm3 CO2.

A wrong: Half the ethene volume.
B wrong: Assumes 1 : 1.
C right: 60.0 cm3.
D wrong: Uses oxygen ratio.

14. Which sample contains the greatest number of molecules?

[Ar: H, 1; C, 12; N, 14; O, 16]

A 1.0 g H2
B 16.0 g CH4
C 14.0 g N2
D 22.0 g CO2

Answer: B

A: H2 Mr = 2, moles = 1.0 / 2 = 0.500 mol molecules
B: CH4 Mr = 16, moles = 16.0 / 16 = 1.00 mol molecules
C: N2 Mr = 28, moles = 14.0 / 28 = 0.500 mol molecules
D: CO2 Mr = 44, moles = 22.0 / 44 = 0.500 mol molecules

A wrong: 0.500 mol molecules.
B right: 1.00 mol molecules, the greatest.
C wrong: 0.500 mol molecules.
D wrong: 0.500 mol molecules.

15. Which sample contains the greatest number of atoms?

A 1 mol He
B 1 mol H2
C 1 mol CO2
D 1 mol CH4

Answer: D

A: 1 mol He atoms = 1 mol atoms
B: 1 mol H2 molecules = 2 mol atoms
C: 1 mol CO2 molecules = 3 mol atoms
D: 1 mol CH4 molecules = 5 mol atoms

A wrong: 1 mol atoms.
B wrong: 2 mol atoms.
C wrong: 3 mol atoms.
D right: 5 mol atoms.

16. How many ions are present in 0.100 mol of sodium sulfate, Na2SO4?

[Avogadro constant = 6.02 x 10^23 mol–1]

A 6.02 x 10^22
B 1.20 x 10^23
C 1.81 x 10^23
D 6.02 x 10^23

Answer: C

Na2SO4 contains 2Na+ and 1SO4 2–, so 3 ions per formula unit.
0.100 mol Na2SO4 gives 0.300 mol ions.
Ions = 0.300 x 6.02 x 10^23 = 1.81 x 10^23

A wrong: Counts only formula units.
B wrong: Counts only two ions per formula unit.
C right: 1.81 x 10^23.
D wrong: Counts 1 mol ions.

17. What mass of sodium chloride is present in 250 cm3 of 0.200 mol/dm3 sodium chloride solution?

[Ar: Na, 23; Cl, 35.5]

A 0.0500 g
B 2.93 g
C 5.85 g
D 11.7 g

Answer: B

250 cm3 = 0.250 dm3
Moles NaCl = 0.200 x 0.250 = 0.0500 mol
Mr NaCl = 58.5
Mass = 0.0500 x 58.5 = 2.93 g

A wrong: Gives moles as grams.
B right: 2.93 g.
C wrong: Uses 0.100 mol.
D wrong: Uses 0.200 mol as if volume were 1 dm3.

18. A solution contains 4.00 g of sodium hydroxide in 250 cm3 of solution.

What is its concentration?

[Ar: Na, 23; O, 16; H, 1]

A 0.100 mol/dm3
B 0.250 mol/dm3
C 0.400 mol/dm3
D 1.60 mol/dm3

Answer: C

Mr NaOH = 40
Moles NaOH = 4.00 / 40 = 0.100 mol
250 cm3 = 0.250 dm3
Concentration = 0.100 / 0.250 = 0.400 mol/dm3

A wrong: Uses volume as 1 dm3.
B wrong: Incorrect division.
C right: 0.400 mol/dm3.
D wrong: Divides the wrong way.

19. 25.0 cm3 of sodium carbonate solution reacts exactly with 20.0 cm3 of 0.100 mol/dm3 hydrochloric acid.

Na2CO3 + 2HCl → 2NaCl + H2O + CO2

What is the concentration of the sodium carbonate solution?

A 0.0200 mol/dm3
B 0.0400 mol/dm3
C 0.0800 mol/dm3
D 0.100 mol/dm3

Answer: B

Moles HCl = 0.100 x 20.0/1000 = 0.00200 mol
Na2CO3 : HCl = 1 : 2
Moles Na2CO3 = 0.00100 mol
Volume Na2CO3 = 25.0 cm3 = 0.0250 dm3
Concentration = 0.00100 / 0.0250 = 0.0400 mol/dm3

A wrong: Too low.
B right: 0.0400 mol/dm3.
C wrong: Uses HCl moles directly.
D wrong: Ignores ratio/volume trap.

20. 10.0 g of impure calcium carbonate produces 1.92 dm3 of carbon dioxide at r.t.p. when reacted with excess acid.

CaCO3 + 2HCl → CaCl2 + H2O + CO2

[molar gas volume at r.t.p. = 24 dm3/mol; Mr of CaCO3 = 100]

What is the percentage purity of the calcium carbonate?

A 20.0%
B 40.0%
C 80.0%
D 96.0%

Answer: C

Moles CO2 = 1.92 / 24 = 0.0800 mol
Moles CaCO3 = 0.0800 mol
Mass pure CaCO3 = 0.0800 x 100 = 8.00 g
Percentage purity = 8.00 / 10.0 x 100 = 80.0%

A wrong: Too low.
B wrong: Half the correct purity.
C right: 80.0%.
D wrong: Uses wrong mole conversion.

Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total A Grades, 11 World Records and 7 Distinctions, Educate A Change.

21. Which statement is correct for gases measured at the same temperature and pressure?

A Equal masses of gases contain equal numbers of molecules.
B Equal volumes of gases contain equal numbers of molecules.
C Equal volumes of gases always have equal masses.
D Equal numbers of molecules always have equal relative molecular masses.

Answer: B

A wrong: Equal masses only contain equal molecules if molar masses are the same.
B right: Equal gas volumes at the same temperature and pressure contain equal numbers of molecules.
C wrong: Equal volumes can have different masses.
D wrong: Molecule number does not determine Mr.

22. A gas occupies 120 cm3 at r.t.p.

How many moles of gas are present?

[molar gas volume at r.t.p. = 24 000 cm3/mol]

A 0.00500 mol
B 0.0500 mol
C 5.00 mol
D 5000 mol

Answer: A

Moles = volume / molar gas volume
= 120 / 24 000
= 0.00500 mol

A right: 0.00500 mol.
B wrong: Ten times too high.
C wrong: Much too high.
D wrong: Total unit disaster.

23. What volume of ammonia gas is formed at r.t.p. when 1.20 dm3 of nitrogen reacts completely with excess hydrogen?

N2 + 3H2 → 2NH3

All gas volumes are measured at the same temperature and pressure.

A 0.600 dm3
B 1.20 dm3
C 2.40 dm3
D 3.60 dm3

Answer: C

N2 : NH3 = 1 : 2
1.20 dm3 N2 forms 2.40 dm3 NH3.

A wrong: Half the nitrogen volume.
B wrong: Assumes 1 : 1.
C right: 2.40 dm3.
D wrong: Uses hydrogen ratio.

24. 60.0 cm3 of hydrogen and 40.0 cm3 of oxygen are sparked together.

2H2 + O2 → 2H2O

What volume of gas remains after cooling to room temperature?

A 0 cm3
B 10.0 cm3 oxygen
C 20.0 cm3 oxygen
D 20.0 cm3 hydrogen

Answer: B

60.0 cm3 H2 needs 30.0 cm3 O2.
O2 available = 40.0 cm3.
Excess O2 = 10.0 cm3.
Water condenses after cooling.

A wrong: Oxygen is in excess.
B right: 10.0 cm3 oxygen remains.
C wrong: Wrong ratio.
D wrong: Hydrogen is fully used.

25. 100 cm3 of carbon monoxide is mixed with 100 cm3 of oxygen and ignited.

2CO + O2 → 2CO2

What is the final volume of gas after reaction, measured at the same temperature and pressure?

A 50 cm3
B 100 cm3
C 150 cm3
D 200 cm3

Answer: C

100 cm3 CO needs 50 cm3 O2.
O2 left = 50 cm3.
CO2 formed = 100 cm3.
Final gas volume = 100 + 50 = 150 cm3.

A wrong: Counts only excess oxygen.
B wrong: Counts only CO2.
C right: 150 cm3.
D wrong: Assumes no reaction change.

26. 0.300 mol of aluminium reacts with excess chlorine.

2Al + 3Cl2 → 2AlCl3

How many moles of chlorine are required?

A 0.200 mol
B 0.300 mol
C 0.450 mol
D 0.600 mol

Answer: C

Al : Cl2 = 2 : 3
Moles Cl2 = 0.300 x 3/2 = 0.450 mol

A wrong: Uses inverse ratio.
B wrong: Assumes 1 : 1.
C right: 0.450 mol.
D wrong: Doubles aluminium moles.

27. Iron reacts with sulfur to form iron sulfide.

Fe + S → FeS

5.60 g of iron reacts with 4.80 g of sulfur.

[Ar: Fe, 56; S, 32]

Which statement is correct?

A Iron is in excess and 8.80 g FeS forms.
B Sulfur is in excess and 8.80 g FeS forms.
C Iron is in excess and 10.4 g FeS forms.
D Sulfur is in excess and 10.4 g FeS forms.

Answer: B

Moles Fe = 5.60 / 56 = 0.100 mol
Moles S = 4.80 / 32 = 0.150 mol
Fe : S = 1 : 1, so iron is limiting and sulfur is excess.
Moles FeS = 0.100 mol
Mr FeS = 88
Mass FeS = 0.100 x 88 = 8.80 g

A wrong: Mass right, excess substance wrong.
B right: Sulfur is in excess and 8.80 g FeS forms.
C wrong: Iron is not in excess and mass is too high.
D wrong: Sulfur excess is right, but mass is too high.

28. Zinc reacts with hydrochloric acid.

Zn + 2HCl → ZnCl2 + H2

What volume of hydrogen is produced at r.t.p. by 6.50 g of zinc reacting with excess acid?

[Ar: Zn, 65; molar gas volume at r.t.p. = 24 dm3/mol]

A 1.20 dm3
B 2.40 dm3
C 4.80 dm3
D 6.50 dm3

Answer: B

Moles Zn = 6.50 / 65 = 0.100 mol
Zn : H2 = 1 : 1
Moles H2 = 0.100 mol
Volume H2 = 0.100 x 24 = 2.40 dm3

A wrong: Half the correct answer.
B right: 2.40 dm3.
C wrong: Doubles the correct answer.
D wrong: Confuses mass with volume.

29. 0.0500 mol of a hydrocarbon CxHy burns to form 0.150 mol of CO2 and 0.200 mol of H2O.

What is the formula of the hydrocarbon?

A C2H4
B C3H4
C C3H8
D C4H8

Answer: C

Carbon atoms per molecule = 0.150 / 0.0500 = 3
Hydrogen atoms per molecule = 2 x 0.200 / 0.0500 = 8
Formula = C3H8

A wrong: Too few carbon and hydrogen atoms.
B wrong: Correct carbon but hydrogen too low.
C right: C3H8.
D wrong: Too many carbon atoms.

30. A compound contains 0.200 mol of carbon atoms, 0.400 mol of hydrogen atoms and 0.100 mol of oxygen atoms.

What is its empirical formula?

A C2H4O
B CH2O
C C2H4O2
D C4H8O2

Answer: A

Ratio C : H : O = 0.200 : 0.400 : 0.100
Divide by 0.100 = 2 : 4 : 1
Empirical formula = C2H4O

A right: C2H4O.
B wrong: Divides carbon incorrectly.
C wrong: Oxygen too high.
D wrong: Not simplest.

Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total A Grades, 11 World Records and 7 Distinctions, Educate A Change.

31. Which sample contains 0.500 mol of particles?

[Ar: H, 1; C, 12; N, 14; O, 16]

A 11.0 g CO2
B 14.0 g N2
C 8.00 g O2
D 18.0 g H2O

Answer: B

A: CO2 Mr = 44, moles = 11.0 / 44 = 0.250 mol
B: N2 Mr = 28, moles = 14.0 / 28 = 0.500 mol
C: O2 Mr = 32, moles = 8.00 / 32 = 0.250 mol
D: H2O Mr = 18, moles = 18.0 / 18 = 1.00 mol

A wrong: 0.250 mol.
B right: 0.500 mol.
C wrong: 0.250 mol.
D wrong: 1.00 mol.

32. How many molecules are present in 4.80 dm3 of oxygen at r.t.p.?

[molar gas volume at r.t.p. = 24 dm3/mol; Avogadro constant = 6.02 x 10^23 mol–1]

A 1.20 x 10^23
B 2.41 x 10^23
C 6.02 x 10^23
D 1.20 x 10^24

Answer: A

Moles O2 = 4.80 / 24 = 0.200 mol
Molecules = 0.200 x 6.02 x 10^23
= 1.20 x 10^23

A right: 1.20 x 10^23.
B wrong: Uses 0.400 mol.
C wrong: Uses 1 mol.
D wrong: Uses 2 mol.

33. What volume of 0.500 mol/dm3 potassium hydroxide is needed to prepare 250 cm3 of 0.100 mol/dm3 potassium hydroxide by dilution?

A 25.0 cm3
B 50.0 cm3
C 100 cm3
D 125 cm3

Answer: B

C1V1 = C2V2
0.500 x V1 = 0.100 x 250
V1 = 50.0 cm3

A wrong: Half the correct volume.
B right: 50.0 cm3.
C wrong: Too large.
D wrong: Uses half the final volume.

34. 25.0 cm3 of 0.200 mol/dm3 hydrochloric acid neutralises 20.0 cm3 of aqueous barium hydroxide.

Ba(OH)2 + 2HCl → BaCl2 + 2H2O

What is the concentration of the barium hydroxide solution?

A 0.0500 mol/dm3
B 0.100 mol/dm3
C 0.125 mol/dm3
D 0.250 mol/dm3

Answer: C

Moles HCl = 0.200 x 25.0/1000 = 0.00500 mol
Ba(OH)2 : HCl = 1 : 2
Moles Ba(OH)2 = 0.00250 mol
Volume Ba(OH)2 = 20.0 cm3 = 0.0200 dm3
Concentration = 0.00250 / 0.0200 = 0.125 mol/dm3

A wrong: Divides by ratio twice.
B wrong: Incorrect ratio/volume handling.
C right: 0.125 mol/dm3.
D wrong: Uses HCl moles directly.

35. 12.0 g of magnesium reacts with excess dilute hydrochloric acid.

Mg + 2HCl → MgCl2 + H2

What volume of hydrogen is produced at r.t.p.?

[Ar: Mg, 24; molar gas volume at r.t.p. = 24 dm3/mol]

A 6.00 dm3
B 12.0 dm3
C 24.0 dm3
D 48.0 dm3

Answer: B

Moles Mg = 12.0 / 24 = 0.500 mol
Mg : H2 = 1 : 1
Moles H2 = 0.500 mol
Volume H2 = 0.500 x 24 = 12.0 dm3

A wrong: Half the correct answer.
B right: 12.0 dm3.
C wrong: Uses 1 mol hydrogen.
D wrong: Doubles again.

36. A student makes 500 cm3 of 0.200 mol/dm3 copper(II) sulfate solution.

What amount of copper(II) sulfate is required?

A 0.0100 mol
B 0.100 mol
C 0.200 mol
D 100 mol

Answer: B

500 cm3 = 0.500 dm3
Moles = concentration x volume
= 0.200 x 0.500
= 0.100 mol

A wrong: Uses 0.0500 dm3.
B right: 0.100 mol.
C wrong: Ignores the 0.500 dm3 volume.
D wrong: Impossible.

37. Which expression correctly calculates the concentration, in mol/dm3, when n moles of solute are dissolved to make V cm3 of solution?

A n / V
B n x V
C n / (V / 1000)
D n x (V / 1000)

Answer: C

V cm3 = V/1000 dm3
Concentration = moles / volume in dm3
= n / (V/1000)

A wrong: Uses cm3 directly.
B wrong: Multiplies instead of dividing.
C right: n / (V / 1000).
D wrong: Multiplies by volume.

38. 0.100 mol of hydrated sodium carbonate, Na2CO3.10H2O, is used.

What mass is used?

[Mr of Na2CO3.10H2O = 286]

A 10.6 g
B 18.0 g
C 28.6 g
D 286 g

Answer: C

Mass = moles x Mr
= 0.100 x 286
= 28.6 g

A wrong: Uses anhydrous sodium carbonate.
B wrong: Uses only water of crystallisation.
C right: 28.6 g.
D wrong: Mass of 1 mol.

39. A student has 0.200 mol of oxygen molecules, O2.

How many moles of oxygen atoms are present?

A 0.100 mol
B 0.200 mol
C 0.400 mol
D 6.02 x 10^23 mol

Answer: C

Each O2 molecule contains 2 oxygen atoms.
0.200 mol O2 molecules contains 0.400 mol oxygen atoms.

A wrong: Halves instead of doubles.
B wrong: Counts molecules, not atoms.
C right: 0.400 mol.
D wrong: Avogadro constant is not a mole amount here.

40. 2.00 dm3 of 0.250 mol/dm3 sodium chloride solution is evaporated to dryness.

What mass of sodium chloride remains?

[Ar: Na, 23; Cl, 35.5]

A 14.6 g
B 29.3 g
C 58.5 g
D 117 g

Answer: B

Moles NaCl = 0.250 x 2.00 = 0.500 mol
Mr NaCl = 58.5
Mass = 0.500 x 58.5 = 29.25 g ≈ 29.3 g

A wrong: Half the correct mass.
B right: 29.3 g.
C wrong: Mass of 1 mol.
D wrong: Mass of 2 mol.

Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total A Grades, 11 World Records and 7 Distinctions, Educate A Change.

41. A student reacts 0.100 mol of calcium with 0.150 mol of chlorine.

Ca + Cl2 → CaCl2

Which statement is correct?

A Calcium is limiting and 0.100 mol CaCl2 forms.
B Chlorine is limiting and 0.100 mol CaCl2 forms.
C Calcium is limiting and 0.150 mol CaCl2 forms.
D Chlorine is limiting and 0.150 mol CaCl2 forms.

Answer: A

Ca : Cl2 = 1 : 1
0.100 mol Ca needs 0.100 mol Cl2.
0.150 mol Cl2 is available, so chlorine is excess and calcium is limiting.
Moles CaCl2 formed = 0.100 mol.

A right: Calcium is limiting and 0.100 mol CaCl2 forms.
B wrong: Chlorine is not limiting.
C wrong: Product is not 0.150 mol.
D wrong: Chlorine is not limiting.

42. 4.80 g of carbon reacts completely with oxygen to form carbon dioxide.

C + O2 → CO2

What volume of carbon dioxide forms at r.t.p.?

[Ar: C, 12; molar gas volume at r.t.p. = 24 dm3/mol]

A 2.40 dm3
B 4.80 dm3
C 9.60 dm3
D 19.2 dm3

Answer: C

Moles C = 4.80 / 12 = 0.400 mol
C : CO2 = 1 : 1
Moles CO2 = 0.400 mol
Volume CO2 = 0.400 x 24 = 9.60 dm3

A wrong: Uses 0.100 mol.
B wrong: Uses 0.200 mol.
C right: 9.60 dm3.
D wrong: Doubles the correct answer.

43. 2.00 g of hydrogen reacts with 16.0 g of oxygen.

2H2 + O2 → 2H2O

[Ar: H, 1; O, 16]

Which mass of water is formed?

A 9.00 g
B 18.0 g
C 20.0 g
D 36.0 g

Answer: B

Moles H2 = 2.00 / 2 = 1.00 mol
Moles O2 = 16.0 / 32 = 0.500 mol
Ratio H2 : O2 = 2 : 1, so both react completely.
Moles H2O formed = 1.00 mol
Mass H2O = 1.00 x 18 = 18.0 g

A wrong: Uses 0.500 mol water.
B right: 18.0 g.
C wrong: Adds reactant masses without mole ratio thinking.
D wrong: Uses 2.00 mol water.

44. 0.0400 mol of an acid H2A reacts exactly with 0.0800 mol of sodium hydroxide.

H2A + 2NaOH → Na2A + 2H2O

What volume of 0.200 mol/dm3 sodium hydroxide contains 0.0800 mol?

A 40.0 cm3
B 80.0 cm3
C 200 cm3
D 400 cm3

Answer: D

Volume = moles / concentration
= 0.0800 / 0.200
= 0.400 dm3
= 400 cm3

A wrong: Contains only 0.00800 mol.
B wrong: Contains only 0.0160 mol.
C wrong: Contains only 0.0400 mol.
D right: 400 cm3.

45. What mass of calcium oxide is formed by heating 5.00 g of calcium carbonate?

CaCO3 → CaO + CO2

[Mr: CaCO3 = 100; CaO = 56]

A 1.40 g
B 2.80 g
C 5.00 g
D 7.14 g

Answer: B

Moles CaCO3 = 5.00 / 100 = 0.0500 mol
CaCO3 : CaO = 1 : 1
Moles CaO = 0.0500 mol
Mass CaO = 0.0500 x 56 = 2.80 g

A wrong: Half the correct mass.
B right: 2.80 g.
C wrong: Assumes product mass equals reactant mass.
D wrong: Uses inverse mass ratio.

46. 0.250 mol of a gas has a mass of 11.0 g.

What is the relative molecular mass of the gas?

A 22.0
B 27.5
C 44.0
D 88.0

Answer: C

Mr = mass / moles
= 11.0 / 0.250
= 44.0

A wrong: Half the correct value.
B wrong: Incorrect division.
C right: 44.0.
D wrong: Doubles the correct value.

47. Which sample has the same number of molecules as 12.0 dm3 of hydrogen at r.t.p.?

[molar gas volume at r.t.p. = 24 dm3/mol]

A 0.250 mol carbon dioxide
B 0.500 mol carbon dioxide
C 1.00 mol carbon dioxide
D 12.0 mol carbon dioxide

Answer: B

12.0 dm3 hydrogen at r.t.p. = 12.0 / 24 = 0.500 mol molecules.
So the matching sample must contain 0.500 mol molecules.

A wrong: Half as many molecules.
B right: Same number of molecules.
C wrong: Twice as many molecules.
D wrong: Far too many.

48. 25.0 cm3 of 0.100 mol/dm3 potassium manganate(VII) solution contains how many moles of KMnO4?

A 0.000250 mol
B 0.00250 mol
C 0.0250 mol
D 2.50 mol

Answer: B

Moles = concentration x volume
= 0.100 x 25.0/1000
= 0.00250 mol

A wrong: Ten times too small.
B right: 0.00250 mol.
C wrong: Does not convert cm3 to dm3.
D wrong: Impossible overestimate.

49. A reaction produces 0.0600 mol of carbon dioxide.

What volume of carbon dioxide is produced at r.t.p.?

[molar gas volume at r.t.p. = 24 dm3/mol]

A 0.144 dm3
B 1.44 dm3
C 2.50 dm3
D 14.4 dm3

Answer: B

Volume = moles x 24
= 0.0600 x 24
= 1.44 dm3

A wrong: Uses 2.4 instead of 24.
B right: 1.44 dm3.
C wrong: Divides wrongly.
D wrong: Ten times too high.

50. 100 cm3 of ethane is completely combusted in 500 cm3 of oxygen.

2C2H6 + 7O2 → 4CO2 + 6H2O

All gas volumes are measured at the same temperature and pressure. What is the final volume of gas after cooling to room temperature?

A 200 cm3
B 300 cm3
C 350 cm3
D 550 cm3

Answer: C

Ethane : oxygen = 2 : 7
100 cm3 ethane needs 350 cm3 oxygen.
Oxygen available = 500 cm3.
Excess oxygen = 150 cm3.

Ethane : carbon dioxide = 2 : 4, so 100 cm3 ethane forms 200 cm3 CO2.
Water condenses after cooling to room temperature.

Final gas volume = CO2 + excess O2
= 200 + 150
= 350 cm3

A wrong: Counts only CO2.
B wrong: Misses 50 cm3 gas.
C right: 350 cm3.
D wrong: Includes too much gas after cooling.

Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total A Grades, 11 World Records and 7 Distinctions, Educate A Change.