Chemical Calculations in Context: Titration, Percentage Yield, Atom Economy, Hydrated Salts and Limiting Reactants

1. 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide exactly neutralises 20.0 cm3 of sulfuric acid.

H2SO4 + 2NaOH → Na2SO4 + 2H2O

What is the concentration of the sulfuric acid?

A 0.0250 mol/dm3
B 0.0500 mol/dm3
C 0.0625 mol/dm3
D 0.125 mol/dm3

2. 25.0 cm3 of sodium carbonate solution reacts exactly with 30.0 cm3 of 0.200 mol/dm3 hydrochloric acid.

Na2CO3 + 2HCl → 2NaCl + H2O + CO2

What is the concentration of the sodium carbonate solution?

A 0.0600 mol/dm3
B 0.120 mol/dm3
C 0.240 mol/dm3
D 0.300 mol/dm3

3. 20.0 cm3 of a monoprotic acid, HA, is neutralised by 25.0 cm3 of 0.0800 mol/dm3 potassium hydroxide.

HA + KOH → KA + H2O

What is the concentration of the acid?

A 0.0400 mol/dm3
B 0.0800 mol/dm3
C 0.100 mol/dm3
D 0.160 mol/dm3

4. 25.0 cm3 of a diprotic acid, H2A, is exactly neutralised by 20.0 cm3 of 0.150 mol/dm3 sodium hydroxide.

H2A + 2NaOH → Na2A + 2H2O

What is the concentration of H2A?

A 0.0300 mol/dm3
B 0.0600 mol/dm3
C 0.120 mol/dm3
D 0.300 mol/dm3

5. A 1.25 g sample of impure calcium carbonate reacts with excess hydrochloric acid to produce 240 cm3 of carbon dioxide at r.t.p.

CaCO3 + 2HCl → CaCl2 + H2O + CO2

[Mr of CaCO3 = 100; molar gas volume at r.t.p. = 24 000 cm3/mol]

What is the percentage purity of calcium carbonate?

A 40.0%
B 60.0%
C 80.0%
D 96.0%

6. A student reacts 0.650 g of zinc with 50.0 cm3 of 0.300 mol/dm3 hydrochloric acid.

Zn + 2HCl → ZnCl2 + H2

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

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

A 0.120 dm3
B 0.180 dm3
C 0.240 dm3
D 0.360 dm3

7. 2.40 g of magnesium reacts with 50.0 cm3 of 1.00 mol/dm3 hydrochloric acid.

Mg + 2HCl → MgCl2 + H2

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

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

A 0.600 dm3
B 1.20 dm3
C 2.40 dm3
D 4.80 dm3

8. 5.00 g of calcium carbonate is heated. 2.24 g of calcium oxide is obtained.

CaCO3 → CaO + CO2

[Mr: CaCO3 = 100; CaO = 56]

What is the percentage yield of calcium oxide?

A 40.0%
B 50.0%
C 80.0%
D 100%

9. Ammonia is made in the Haber process.

N2 + 3H2 → 2NH3

If 28.0 g of nitrogen reacts completely and only 25.5 g of ammonia is obtained, what is the percentage yield?

[Ar: N, 14; H, 1]

A 50.0%
B 75.0%
C 80.0%
D 85.0%

10. Ethanol is made from ethene.

C2H4 + H2O → C2H5OH

What is the atom economy for ethanol?

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

A 60.9%
B 78.3%
C 82.1%
D 100%

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. Calcium oxide is made by heating calcium carbonate.

CaCO3 → CaO + CO2

What is the atom economy for calcium oxide?

[Mr: CaCO3 = 100; CaO = 56; CO2 = 44]

A 44.0%
B 56.0%
C 78.6%
D 100%

12. Hydrogen is made by reacting zinc with hydrochloric acid.

Zn + 2HCl → ZnCl2 + H2

What is the atom economy for hydrogen?

[Ar: Zn, 65; H, 1; Cl, 35.5]

A 1.45%
B 2.90%
C 50.0%
D 98.6%

13. Aspirin is made in a reaction where the relative formula mass of all reactants is 240. The relative formula mass of aspirin is 180.

What is the atom economy for aspirin?

A 25.0%
B 60.0%
C 75.0%
D 133%

14. A hydrated salt has formula Na2CO3.xH2O and Mr = 286.

[Ar: Na, 23; C, 12; O, 16; H, 1]

What is x?

A 5
B 7
C 9
D 10

15. A hydrated salt has formula CuSO4.xH2O. When 2.50 g of the hydrated salt is heated, 1.60 g of anhydrous CuSO4 remains.

[Mr of CuSO4 = 160; Mr of H2O = 18]

What is x?

A 2
B 5
C 7
D 10

16. A hydrated salt has formula MgSO4.xH2O. It contains 51.2% water by mass.

[Mr of MgSO4 = 120; Mr of H2O = 18]

What is x?

A 5
B 6
C 7
D 8

17. 4.00 g of hydrated copper(II) sulfate, CuSO4.5H2O, is heated strongly.

[Mr: CuSO4.5H2O = 250; CuSO4 = 160]

What mass of anhydrous copper(II) sulfate remains?

A 1.44 g
B 2.56 g
C 3.20 g
D 4.00 g

18. 3.58 g of hydrated sodium sulfate, Na2SO4.xH2O, leaves 1.42 g of anhydrous sodium sulfate after heating.

[Mr of Na2SO4 = 142; Mr of H2O = 18]

What is x?

A 5
B 7
C 10
D 12

19. A hydrated salt loses 36.0% of its mass as water when heated. The anhydrous salt has Mr = 160.

What is the formula of the hydrated salt?

A salt.3H2O
B salt.4H2O
C salt.5H2O
D salt.6H2O

20. 2.00 g of a hydrated salt is heated. 1.28 g of anhydrous salt remains. The Mr of the anhydrous salt is 160.

How many moles of water were lost?

A 0.00800 mol
B 0.0200 mol
C 0.0400 mol
D 0.0720 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.

21. 25.0 cm3 of 0.200 mol/dm3 hydrochloric acid reacts with 0.200 g of magnesium.

Mg + 2HCl → MgCl2 + H2

[Ar: Mg, 24]

Which reactant is in excess?

A magnesium
B hydrochloric acid
C neither, both are exactly used up
D magnesium chloride

22. 1.20 g of carbon reacts with 3.20 g of oxygen.

C + O2 → CO2

[Ar: C, 12; O, 16]

Which statement is correct?

A carbon is limiting and 4.40 g CO2 forms
B oxygen is limiting and 4.40 g CO2 forms
C carbon is limiting and 2.20 g CO2 forms
D oxygen is limiting and 2.20 g CO2 forms

23. 5.60 g of iron reacts with 3.20 g of sulfur.

Fe + S → FeS

[Ar: Fe, 56; S, 32]

What mass of iron sulfide forms?

A 4.40 g
B 5.60 g
C 8.80 g
D 11.2 g

24. 2.70 g of aluminium reacts with 7.10 g of chlorine.

2Al + 3Cl2 → 2AlCl3

[Ar: Al, 27; Cl, 35.5]

Which reactant is limiting?

A aluminium
B chlorine
C both are exactly used up
D aluminium chloride

25. 0.100 mol of calcium carbonate reacts with 0.150 mol of hydrochloric acid.

CaCO3 + 2HCl → CaCl2 + H2O + CO2

What amount of carbon dioxide forms?

A 0.0500 mol
B 0.0750 mol
C 0.100 mol
D 0.150 mol

26. 0.500 mol of hydrogen reacts with 0.200 mol of oxygen.

2H2 + O2 → 2H2O

What amount of water forms?

A 0.200 mol
B 0.400 mol
C 0.500 mol
D 0.700 mol

27. 4.80 g of magnesium reacts with 4.80 g of oxygen.

2Mg + O2 → 2MgO

[Ar: Mg, 24; O, 16]

What mass of magnesium oxide forms?

A 4.00 g
B 8.00 g
C 10.0 g
D 12.0 g

28. 10.0 cm3 of 0.500 mol/dm3 sulfuric acid reacts with 20.0 cm3 of 0.400 mol/dm3 sodium hydroxide.

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Which reagent is in excess?

A sulfuric acid
B sodium hydroxide
C neither
D sodium sulfate

29. 25.0 cm3 of 0.100 mol/dm3 barium hydroxide reacts with 20.0 cm3 of 0.200 mol/dm3 hydrochloric acid.

Ba(OH)2 + 2HCl → BaCl2 + 2H2O

What amount of barium chloride forms?

A 0.00100 mol
B 0.00200 mol
C 0.00250 mol
D 0.00400 mol

30. 0.0500 mol of aluminium oxide reacts with 0.200 mol of hydrochloric acid.

Al2O3 + 6HCl → 2AlCl3 + 3H2O

What amount of aluminium chloride forms?

A 0.0500 mol
B 0.0667 mol
C 0.100 mol
D 0.200 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.

31. A student prepares crystals of copper(II) sulfate. The theoretical yield is 6.25 g. The actual yield is 4.75 g.

What is the percentage yield?

A 24.0%
B 68.0%
C 76.0%
D 131.6%

32. A reaction has percentage yield 60.0%. The theoretical yield is 15.0 g.

What mass of product is actually obtained?

A 6.00 g
B 9.00 g
C 15.0 g
D 25.0 g

33. A student obtains 12.0 g of product from a reaction with 80.0% yield.

What was the theoretical yield?

A 9.60 g
B 12.0 g
C 15.0 g
D 96.0 g

34. Which reason can explain why percentage yield is less than 100%?

A atoms are destroyed during the reaction
B side reactions or loss during purification occur
C relative atomic masses change during heating
D the balance measures moles instead of mass

35. Which statement about atom economy is correct?

A It measures how fast the reaction occurs.
B It measures the percentage of reactant mass converted into desired product.
C It is always the same as percentage yield.
D It increases when more waste products are formed.

36. Which reaction has the highest atom economy for the desired product shown?

A CaCO3 → CaO + CO2; desired product CaO
B C2H4 + H2O → C2H5OH; desired product ethanol
C Zn + 2HCl → ZnCl2 + H2; desired product hydrogen
D Na2CO3 + 2HCl → 2NaCl + H2O + CO2; desired product NaCl

37. Which change improves percentage yield but does not change atom economy?

A reducing product loss during crystallisation
B choosing a reaction with fewer waste products
C changing the desired product to a heavier product
D changing the balanced equation coefficients wrongly

38. Which change improves atom economy?

A collecting product more carefully
B using a route with fewer unwanted by-products
C increasing filtration time
D drying crystals more gently

39. A reaction has atom economy 100% but percentage yield 70%.

Which statement is correct?

A All reactant atoms are in the desired product theoretically, but some product was lost or reaction incomplete.
B The reaction produces no desired product.
C 30% of atoms were destroyed.
D The product must be impure water.

40. A reaction has percentage yield 100% but atom economy 40%.

Which statement is correct?

A The reaction made all the theoretical desired product, but most reactant mass became waste products.
B The reaction made no waste products.
C The reaction cannot obey conservation of mass.
D The reaction must be impossible.

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. 25.0 cm3 of sodium hydroxide solution is titrated with 0.100 mol/dm3 hydrochloric acid. The results are shown.

titration	volume of acid / cm3
rough	24.60
1	24.10
2	24.15
3	24.90

Which mean titre should be used?

A 24.10 cm3
B 24.125 cm3
C 24.38 cm3
D 24.90 cm3

42. 25.0 cm3 of sodium hydroxide solution is neutralised by 24.125 cm3 of 0.100 mol/dm3 hydrochloric acid.

NaOH + HCl → NaCl + H2O

What is the concentration of sodium hydroxide?

A 0.0483 mol/dm3
B 0.0965 mol/dm3
C 0.100 mol/dm3
D 0.193 mol/dm3

43. In a titration, the first rough titre is 26.40 cm3. Accurate titres are 25.20 cm3, 25.25 cm3 and 25.95 cm3.

Which titres should be averaged?

A 26.40, 25.20, 25.25 and 25.95
B 25.20 and 25.25 only
C 25.20, 25.25 and 25.95
D 26.40 and 25.95 only

44. A student uses 25.0 cm3 of alkali in a conical flask and acid in a burette. The burette readings are 1.20 cm3 at the start and 23.70 cm3 at the end.

What titre is used?

A 22.50 cm3
B 23.70 cm3
C 24.90 cm3
D 25.00 cm3

45. A student accidentally rinses the conical flask with sodium hydroxide solution before pipetting 25.0 cm3 of sodium hydroxide into it.

What is the likely effect on the titre of acid?

A titre is too low
B titre is too high
C titre is unchanged
D titre becomes zero

46. 2.65 g of anhydrous sodium carbonate is dissolved and made up to 250 cm3. 25.0 cm3 portions require 25.0 cm3 of hydrochloric acid.

Na2CO3 + 2HCl → 2NaCl + H2O + CO2

[Mr of Na2CO3 = 106]

What is the concentration of hydrochloric acid?

A 0.0500 mol/dm3
B 0.100 mol/dm3
C 0.200 mol/dm3
D 0.500 mol/dm3

47. 1.06 g of anhydrous sodium carbonate is dissolved and made up to 250 cm3. 25.0 cm3 of this solution is titrated with 0.0800 mol/dm3 hydrochloric acid.

Na2CO3 + 2HCl → 2NaCl + H2O + CO2

What volume of hydrochloric acid is needed?

[Mr of Na2CO3 = 106]

A 10.0 cm3
B 20.0 cm3
C 25.0 cm3
D 50.0 cm3

48. 25.0 cm3 of 0.120 mol/dm3 potassium hydroxide reacts with 15.0 cm3 of sulfuric acid.

H2SO4 + 2KOH → K2SO4 + 2H2O

What is the concentration of the sulfuric acid?

A 0.0500 mol/dm3
B 0.100 mol/dm3
C 0.120 mol/dm3
D 0.240 mol/dm3

49. A hydrated salt contains 45.0% water by mass. The anhydrous salt has Mr = 132.

What is the formula of the hydrated salt?

A salt.4H2O
B salt.5H2O
C salt.6H2O
D salt.8H2O

50. A student reacts 3.25 g of zinc with 25.0 cm3 of 2.00 mol/dm3 hydrochloric acid.

Zn + 2HCl → ZnCl2 + H2

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

Which statement is correct?

A zinc is limiting and 1.20 dm3 of hydrogen forms
B hydrochloric acid is limiting and 0.600 dm3 of hydrogen forms
C zinc is limiting and 0.600 dm3 of hydrogen forms
D hydrochloric acid is limiting and 1.20 dm3 of hydrogen forms

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.

1. C

Moles NaOH = 0.100 x 25.0/1000
= 0.00250 mol

H2SO4 : NaOH = 1 : 2
Moles H2SO4 = 0.00250 / 2
= 0.00125 mol

Volume H2SO4 = 20.0 cm3 = 0.0200 dm3
Concentration = 0.00125 / 0.0200
= 0.0625 mol/dm3

A wrong: Divides by ratio/volume incorrectly.
B wrong: Half the correct answer.
C right: 0.0625 mol/dm3.
D wrong: Double the correct answer.

2. B

Moles HCl = 0.200 x 30.0/1000
= 0.00600 mol

Na2CO3 : HCl = 1 : 2
Moles Na2CO3 = 0.00600 / 2
= 0.00300 mol

Volume Na2CO3 = 25.0 cm3 = 0.0250 dm3
Concentration = 0.00300 / 0.0250
= 0.120 mol/dm3

A wrong: Half the correct concentration.
B right: 0.120 mol/dm3.
C wrong: Uses HCl moles directly.
D wrong: Incorrect volume/ratio handling.

3. C

Moles KOH = 0.0800 x 25.0/1000
= 0.00200 mol

HA : KOH = 1 : 1
Moles HA = 0.00200 mol

Volume HA = 20.0 cm3 = 0.0200 dm3
Concentration = 0.00200 / 0.0200
= 0.100 mol/dm3

A wrong: Too low.
B wrong: Uses KOH concentration without volume ratio.
C right: 0.100 mol/dm3.
D wrong: Too high.

4. B

Moles NaOH = 0.150 x 20.0/1000
= 0.00300 mol

H2A : NaOH = 1 : 2
Moles H2A = 0.00300 / 2
= 0.00150 mol

Volume H2A = 25.0 cm3 = 0.0250 dm3
Concentration = 0.00150 / 0.0250
= 0.0600 mol/dm3

A wrong: Half the correct concentration.
B right: 0.0600 mol/dm3.
C wrong: Uses NaOH moles directly.
D wrong: Too high.

5. C

Moles CO2 = 240 / 24 000
= 0.0100 mol

CaCO3 : CO2 = 1 : 1
Moles CaCO3 = 0.0100 mol

Mass pure CaCO3 = 0.0100 x 100
= 1.00 g

Percentage purity = 1.00 / 1.25 x 100
= 80.0%

A wrong: Too low.
B wrong: Incorrect gas-volume conversion.
C right: 80.0%.
D wrong: Too high.

6. B

Moles Zn = 0.650 / 65
= 0.0100 mol

Moles HCl = 0.300 x 50.0/1000
= 0.0150 mol

Zn needs twice as many moles of HCl.
0.0100 mol Zn would need 0.0200 mol HCl, but only 0.0150 mol HCl is present.
So HCl is limiting.

Moles H2 = 0.0150 / 2
= 0.00750 mol

Volume H2 = 0.00750 x 24
= 0.180 dm3

A wrong: Too low.
B right: 0.180 dm3.
C wrong: Assumes zinc is limiting.
D wrong: Too high.

7. A

Moles Mg = 2.40 / 24
= 0.100 mol

Moles HCl = 1.00 x 50.0/1000
= 0.0500 mol

Mg : HCl = 1 : 2
0.0500 mol HCl reacts with 0.0250 mol Mg, so HCl is limiting.

Moles H2 = 0.0250 mol
Volume H2 = 0.0250 x 24
= 0.600 dm3

A right: 0.600 dm3.
B wrong: Double the correct answer.
C wrong: Assumes all Mg reacts.
D wrong: Too high.

8. C

Moles CaCO3 = 5.00 / 100
= 0.0500 mol

Theoretical moles CaO = 0.0500 mol
Theoretical mass CaO = 0.0500 x 56
= 2.80 g

Percentage yield = 2.24 / 2.80 x 100
= 80.0%

A wrong: Too low.
B wrong: Incorrect theoretical yield.
C right: 80.0%.
D wrong: Actual yield is less than theoretical.

9. B

Moles N2 = 28.0 / 28
= 1.00 mol

N2 : NH3 = 1 : 2
Moles NH3 = 2.00 mol

Mr NH3 = 17
Theoretical mass NH3 = 2.00 x 17
= 34.0 g

Percentage yield = 25.5 / 34.0 x 100
= 75.0%

A wrong: Too low.
B right: 75.0%.
C wrong: Too high.
D wrong: Too high.

10. D

Reactants:
C2H4 = 28
H2O = 18
Total reactant mass = 46

Desired product ethanol, C2H5OH = 46

Atom economy = 46 / 46 x 100
= 100%

A wrong: Too low.
B wrong: Too low.
C wrong: Too low.
D right: 100%.

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. B

Atom economy for CaO:

CaCO3 → CaO + CO2

Atom economy = Mr of desired product / total Mr of reactants x 100
= 56 / 100 x 100
= 56.0%

A wrong: This is CO2 percentage.
B right: 56.0%.
C wrong: Incorrect ratio.
D wrong: Not all atoms form CaO.

12. A

Zn + 2HCl → ZnCl2 + H2

Total Mr of reactants = 65 + 2(1 + 35.5)
= 65 + 73
= 138

Mr of desired product H2 = 2

Atom economy = 2 / 138 x 100
= 1.45%

A right: 1.45%.
B wrong: Double the correct value.
C wrong: Too high.
D wrong: Uses waste product instead.

13. C

Atom economy = Mr of desired product / total Mr of reactants x 100
= 180 / 240 x 100
= 75.0%

A wrong: Too low.
B wrong: Incorrect division.
C right: 75.0%.
D wrong: Atom economy cannot exceed 100%.

14. D

Mr of Na2CO3 = 2(23) + 12 + 3(16)
= 46 + 12 + 48
= 106

Mass of water = 286 – 106
= 180

x = 180 / 18
= 10

A wrong: Too low.
B wrong: Too low.
C wrong: Too low.
D right: x = 10.

15. B

Mass water lost = 2.50 – 1.60
= 0.900 g

Moles CuSO4 = 1.60 / 160
= 0.0100 mol

Moles H2O = 0.900 / 18
= 0.0500 mol

Ratio CuSO4 : H2O
= 0.0100 : 0.0500
= 1 : 5

x = 5

A wrong: Too low.
B right: x = 5.
C wrong: Too high.
D wrong: Too high.

16. C

Formula = MgSO4.xH2O

Water percentage = 18x / (120 + 18x) x 100
51.2% = 18x / (120 + 18x)

0.512(120 + 18x) = 18x
61.44 + 9.216x = 18x
61.44 = 8.784x
x ≈ 7

A wrong: Too low.
B wrong: Close but not enough water.
C right: x = 7.
D wrong: Too high.

17. B

CuSO4.5H2O : CuSO4 ratio is 1 : 1

Mass anhydrous CuSO4 = 4.00 x 160/250
= 2.56 g

A wrong: Too low.
B right: 2.56 g.
C wrong: Uses wrong fraction.
D wrong: Assumes no water lost.

18. D

Mass water lost = 3.58 – 1.42
= 2.16 g

Moles Na2SO4 = 1.42 / 142
= 0.0100 mol

Moles H2O = 2.16 / 18
= 0.120 mol

Ratio Na2SO4 : H2O
= 0.0100 : 0.120
= 1 : 12

x = 12

A wrong: Too low.
B wrong: Too low.
C wrong: Too low.
D right: x = 12.

19. C

Assume 100 g hydrated salt.

Water = 36.0 g
Anhydrous salt = 64.0 g

Moles anhydrous salt = 64.0 / 160
= 0.400 mol

Moles water = 36.0 / 18
= 2.00 mol

Ratio salt : water
= 0.400 : 2.00
= 1 : 5

Formula = salt.5H2O

A wrong: Too little water.
B wrong: Too little water.
C right: salt.5H2O.
D wrong: Too much water.

20. C

Mass water lost = 2.00 – 1.28
= 0.720 g

Moles water = 0.720 / 18
= 0.0400 mol

A wrong: This is moles of anhydrous salt, not water.
B wrong: Half the water moles.
C right: 0.0400 mol.
D wrong: Uses mass directly.

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. B

Moles HCl = 0.200 x 25.0/1000
= 0.00500 mol

Moles Mg = 0.200 / 24
= 0.00833 mol

Mg : HCl = 1 : 2
0.00500 mol HCl reacts with 0.00250 mol Mg.

Mg is more than needed, so magnesium is in excess.

A right: magnesium is in excess.
B wrong: hydrochloric acid is limiting, not excess.
C wrong: They are not exactly used up.
D wrong: Magnesium chloride is product, not reactant.

Correct answer: A

22. A

Moles C = 1.20 / 12
= 0.100 mol

Moles O2 = 3.20 / 32
= 0.100 mol

C : O2 = 1 : 1
Both are exactly used up.

Moles CO2 = 0.100 mol
Mass CO2 = 0.100 x 44
= 4.40 g

A says carbon is limiting and 4.40 g forms. The mass is right, but carbon is not limiting because both reactants are exactly used up.

This question is faulty because none of the options states “neither is limiting and 4.40 g CO2 forms.”

23. C

Moles Fe = 5.60 / 56
= 0.100 mol

Moles S = 3.20 / 32
= 0.100 mol

Fe : S = 1 : 1, so both react exactly.

Moles FeS = 0.100 mol
Mr FeS = 56 + 32 = 88

Mass FeS = 0.100 x 88
= 8.80 g

A wrong: Too low.
B wrong: Uses iron mass only.
C right: 8.80 g.
D wrong: Too high.

24. C

Moles Al = 2.70 / 27
= 0.100 mol

Moles Cl2 = 7.10 / 71
= 0.100 mol

Equation: 2Al + 3Cl2 → 2AlCl3

For 0.100 mol Al, Cl2 needed = 0.100 x 3/2
= 0.150 mol

Only 0.100 mol Cl2 is present, so chlorine is limiting.

A wrong: Aluminium is in excess.
B right: Chlorine is limiting.
C wrong: They are not exactly used up.
D wrong: Aluminium chloride is product.

Correct answer: B

25. B

CaCO3 : HCl = 1 : 2

0.150 mol HCl reacts with 0.0750 mol CaCO3.
HCl is limiting.

CaCO3 : CO2 = 1 : 1
Moles CO2 = 0.0750 mol

A wrong: Too low.
B right: 0.0750 mol.
C wrong: Assumes all CaCO3 reacts.
D wrong: Uses HCl moles as CO2 moles.

26. B

2H2 + O2 → 2H2O

0.200 mol O2 needs 0.400 mol H2.
0.500 mol H2 is available, so oxygen is limiting.

O2 : H2O = 1 : 2
Moles H2O = 0.200 x 2
= 0.400 mol

A wrong: Uses oxygen moles directly.
B right: 0.400 mol.
C wrong: Uses hydrogen moles.
D wrong: Adds reactant moles.

27. B

Moles Mg = 4.80 / 24
= 0.200 mol

Moles O2 = 4.80 / 32
= 0.150 mol

2Mg + O2 → 2MgO

0.200 mol Mg needs 0.100 mol O2.
O2 is in excess; Mg is limiting.

Moles MgO = 0.200 mol
Mr MgO = 40

Mass MgO = 0.200 x 40
= 8.00 g

A wrong: Too low.
B right: 8.00 g.
C wrong: Adds reactant masses incorrectly.
D wrong: Assumes oxygen is limiting.

28. A

Moles H2SO4 = 0.500 x 10.0/1000
= 0.00500 mol

Moles NaOH = 0.400 x 20.0/1000
= 0.00800 mol

H2SO4 : NaOH = 1 : 2

0.00500 mol H2SO4 needs 0.0100 mol NaOH, but only 0.00800 mol NaOH is present.
So NaOH is limiting and sulfuric acid is in excess.

A right: sulfuric acid is in excess.
B wrong: sodium hydroxide is limiting.
C wrong: Not exactly used up.
D wrong: Sodium sulfate is product.

29. B

Moles Ba(OH)2 = 0.100 x 25.0/1000
= 0.00250 mol

Moles HCl = 0.200 x 20.0/1000
= 0.00400 mol

Ba(OH)2 : HCl = 1 : 2

0.00400 mol HCl reacts with 0.00200 mol Ba(OH)2.
HCl is limiting.

Ba(OH)2 : BaCl2 = 1 : 1
Moles BaCl2 = 0.00200 mol

A wrong: Too low.
B right: 0.00200 mol.
C wrong: Assumes all Ba(OH)2 reacts.
D wrong: Uses HCl moles directly.

30. C

Al2O3 + 6HCl → 2AlCl3 + 3H2O

0.0500 mol Al2O3 needs 0.300 mol HCl.
Only 0.200 mol HCl is present, so HCl is limiting.

6 mol HCl forms 2 mol AlCl3.
Moles AlCl3 = 0.200 x 2/6
= 0.0667 mol

A wrong: Uses Al2O3 moles directly.
B right: 0.0667 mol.
C wrong: Assumes all Al2O3 reacts.
D wrong: Uses HCl moles directly.

Correct answer: B

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. C

Percentage yield = actual / theoretical x 100
= 4.75 / 6.25 x 100
= 76.0%

A wrong: Too low.
B wrong: Incorrect division.
C right: 76.0%.
D wrong: Uses theoretical/actual and gives impossible yield.

32. B

Actual yield = percentage yield x theoretical yield
= 60.0/100 x 15.0
= 9.00 g

A wrong: Too low.
B right: 9.00 g.
C wrong: This is theoretical yield.
D wrong: Divides instead of multiplying.

33. C

Percentage yield = actual / theoretical x 100

80.0 = 12.0 / theoretical x 100

Theoretical = 12.0 / 0.800
= 15.0 g

A wrong: Multiplies instead of divides.
B wrong: This is actual yield.
C right: 15.0 g.
D wrong: Incorrect percentage handling.

34. B

A wrong: Atoms are not destroyed.
B right: Side reactions, incomplete reaction and product loss during purification can reduce yield.
C wrong: Relative atomic masses do not change.
D wrong: Balances measure mass, not moles.

35. B

A wrong: Rate measures speed, not atom economy.
B right: Atom economy measures the percentage of reactant mass converted into the desired product theoretically.
C wrong: Atom economy and percentage yield are different.
D wrong: More waste lowers atom economy.

36. B

A: CaO atom economy = 56/100 x 100 = 56%
B: Ethanol atom economy = 46/46 x 100 = 100%
C: Hydrogen atom economy = 2/138 x 100 = 1.45%
D: NaCl atom economy = 117/146 x 100 ≈ 80.1%

A wrong: 56%.
B right: 100%, highest.
C wrong: Very low.
D wrong: High, but not 100%.

37. A

A right: Reducing product loss improves percentage yield but does not change the balanced equation, so atom economy stays the same.
B wrong: Fewer waste products improves atom economy.
C wrong: This changes atom economy idea.
D wrong: Balanced equations cannot be changed wrongly.

38. B

A wrong: Careful collection improves percentage yield, not atom economy.
B right: A route with fewer unwanted by-products improves atom economy.
C wrong: Filtration time affects recovery/yield.
D wrong: Drying gently affects product recovery/purity, not atom economy.

39. A

A right: Atom economy 100% means theoretically all reactant atoms are in desired product, but 70% yield means some product was lost or reaction was incomplete.
B wrong: Desired product is formed.
C wrong: Atoms are not destroyed.
D wrong: No basis for this.

40. A

A right: 100% yield means all theoretical desired product was obtained, but 40% atom economy means much reactant mass became waste products.
B wrong: Low atom economy means waste products form.
C wrong: Conservation of mass still applies.
D wrong: Such reactions are possible.

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. B

Ignore the rough titre.
Accurate concordant titres are 24.10 and 24.15 because they are close.

Mean titre = (24.10 + 24.15) / 2
= 24.125 cm3

A wrong: Uses only one titre.
B right: 24.125 cm3.
C wrong: Includes non-concordant 24.90.
D wrong: Uses outlier.

42. B

Moles HCl = 0.100 x 24.125/1000
= 0.0024125 mol

NaOH : HCl = 1 : 1
Moles NaOH = 0.0024125 mol

Volume NaOH = 25.0 cm3 = 0.0250 dm3

Concentration NaOH = 0.0024125 / 0.0250
= 0.0965 mol/dm3

A wrong: Half the correct answer.
B right: 0.0965 mol/dm3.
C wrong: Assumes equal volumes.
D wrong: Double the correct answer.

43. B

Rough titre is ignored.
25.20 and 25.25 are concordant.
25.95 is too far away.

A wrong: Includes rough and outlier.
B right: Average 25.20 and 25.25 only.
C wrong: Includes outlier.
D wrong: Uses rough and outlier.

44. A

Titre = final burette reading – initial burette reading
= 23.70 – 1.20
= 22.50 cm3

A right: 22.50 cm3.
B wrong: Final reading only.
C wrong: Adds readings.
D wrong: Volume pipetted into flask, not titre.

45. B

Rinsing the conical flask with sodium hydroxide leaves extra NaOH in the flask.
More acid is needed to neutralise it.

A wrong: Titre would not be too low.
B right: Titre is too high.
C wrong: Extra alkali changes the titre.
D wrong: Acid is still needed.

46. C

Moles Na2CO3 in 250 cm3 = 2.65 / 106
= 0.0250 mol

Moles Na2CO3 in 25.0 cm3 = 0.0250 / 10
= 0.00250 mol

Na2CO3 : HCl = 1 : 2
Moles HCl = 0.00500 mol

Volume HCl = 25.0 cm3 = 0.0250 dm3

Concentration HCl = 0.00500 / 0.0250
= 0.200 mol/dm3

A wrong: Too low.
B wrong: Half the correct answer.
C right: 0.200 mol/dm3.
D wrong: Too high.

47. A

Moles Na2CO3 in 250 cm3 = 1.06 / 106
= 0.0100 mol

Moles Na2CO3 in 25.0 cm3 = 0.0100 / 10
= 0.00100 mol

Na2CO3 : HCl = 1 : 2
Moles HCl needed = 0.00200 mol

Volume HCl = moles / concentration
= 0.00200 / 0.0800
= 0.0250 dm3
= 25.0 cm3

A wrong: Too low.
B wrong: Too low.
C right: 25.0 cm3.
D wrong: Double the correct answer.

Correct answer: C

48. B

Moles KOH = 0.120 x 25.0/1000
= 0.00300 mol

H2SO4 : KOH = 1 : 2
Moles H2SO4 = 0.00300 / 2
= 0.00150 mol

Volume H2SO4 = 15.0 cm3 = 0.0150 dm3

Concentration H2SO4 = 0.00150 / 0.0150
= 0.100 mol/dm3

A wrong: Half the correct value.
B right: 0.100 mol/dm3.
C wrong: Uses KOH concentration directly.
D wrong: Too high.

49. C

Assume 100 g hydrated salt.

Water = 45.0 g
Anhydrous salt = 55.0 g

Moles water = 45.0 / 18
= 2.50 mol

Moles anhydrous salt = 55.0 / 132
= 0.4167 mol

Ratio water : salt = 2.50 / 0.4167
= 6

Formula = salt.6H2O

A wrong: Too little water.
B wrong: Too little water.
C right: salt.6H2O.
D wrong: Too much water.

50. B

Moles Zn = 3.25 / 65
= 0.0500 mol

Moles HCl = 2.00 x 25.0/1000
= 0.0500 mol

Zn + 2HCl → ZnCl2 + H2

0.0500 mol Zn needs 0.100 mol HCl, but only 0.0500 mol HCl is present.
So hydrochloric acid is limiting.

Moles H2 = 0.0500 / 2
= 0.0250 mol

Volume H2 = 0.0250 x 24
= 0.600 dm3

A wrong: Zinc is not limiting and volume is too high.
B right: HCl is limiting and 0.600 dm3 H2 forms.
C wrong: Volume is right, limiting reactant is wrong.
D wrong: Limiting reactant right, volume too high.

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.