Moles-limiting reagents

Key Vocabulary

• Mole – the amount of a substance; measured in moles (mol).

• Relative Atomic Mass – the mass of one mole of an element’s atoms, found on the periodic table.

• Formula Mass – the total of all atomic masses for the atoms shown in a chemical formula.

• Molecular Mass – another name for formula mass, specifically when referring to molecules.

• Limiting Reagent – the reactant that is present in fewer moles than needed based on the balanced chemical equation, thus limits the reaction.

• Reagent in Excess – the reactant that is available in greater quantity than needed by the stoichiometric ratio in the reaction.

Limiting Reactants

• In chemical reactions, the amount of product formed often depends on one specific reactant (the other is usually said to be in excess).

• You are expected to determine the limiting reactant yourself.

Definitions

• Reagent – a reactant involved in the chemical reaction.
• Limiting Reagent – the reactant that restricts how much product can be formed; once it is used up, no more product is made.
• In Excess – the reactant that is present in greater quantity than needed and will remain after the reaction finishes.

Method

Step 1 – Write a balanced chemical equation if one is not already provided.
Step 2 – Calculate the number of moles for each reagent available.
Step 3 – Use the mole ratio from the balanced equation to find the required moles of the other reagent.
Step 4 – Compare available and required moles:

• If available moles < required moles → it is the limiting reagent.

• If available moles > required moles → it is the reagent in excess. Step 5 – Use the amount of the limiting reagent to calculate the maximum amount of product formed.

Example 1:

Iron forms when iron(III) oxide is heated with carbon. The equation for the reaction is:
Fe₂O₃ + 3C → 2Fe + 3CO

If you had 0.25 moles of iron(III) oxide and 0.5 moles of carbon, determine which is in excess.

• 0.5 moles of carbon available, but 0.75 moles needed.

• Therefore, carbon is limiting, and Fe₂O₃ is in excess.

Example 2:

10 g of aluminium is reacted with 10 g of oxygen. Identify the limiting reagent and which one is in excess. What is the maximum mass of aluminium oxide that can be produced from this mixture?

4Al + 3O₂ → 2Al₂O₃

• 0.313 moles of O₂ available, 0.278 moles needed.

• O₂ is in excess, so Al is limiting.

Maximum mass of Al₂O₃:

• Moles of Al₂O₃ = 0.370 × (2/4) = 0.185

• Mr of Al₂O₃ = (2 × 27) + (3 × 16) = 102

• Mass = 0.185 × 102 = 18.87 g

Example 3

10 g of nitrogen is reacted with 10 g of hydrogen. Identify the limiting reagent and which one is in excess. What is the maximum mass of ammonia that can be produced from this mixture? If 10 g of ammonia is obtained, what yield does this represent?

N₂ + 3H₂ → 2NH₃

• 5.00 moles of H₂ available, but only 1.071 moles needed.

• Therefore, N₂ is limiting, and H₂ is in excess.

Maximum mass of NH₃:

• Moles of NH₃ = 0.357 × (2/1) = 0.714

• Mr of NH₃ = (14) + (3 × 1) = 17

• Mass = 0.714 × 17 = 12.14 g

Percentage Yield:

• Actual mass = 10 g

• Percentage yield = (10 ÷ 12.14) × 100 = 82.4%

Questions

1. Take the reaction: NH₃ + O₂ → NO + H₂O.
In an experiment, 3.25 g of NH₃ are allowed to react with 3.50 g of O₂.

a) Which reactant is the limiting reagent?

• 0.109 moles of O₂ are available but 0.191 moles are needed.

• Therefore, O₂ is the limiting reagent.

b) How many grams of NO are formed?

• Moles of NO formed = moles of limiting reagent = 0.109

• Mr of NO = 14 + 16 = 30

• Mass = 0.109 × 30 = 3.27 g

c) How much of the excess reactant remains after the reaction?

• Moles of NH₃ needed = moles of O₂ (1:1 ratio) = 0.109

• Moles of NH₃ used = 0.109

• Moles of NH₃ remaining = 0.191 − 0.109 = 0.082

• Mass remaining = 0.082 × 17 = 1.39 g

2. If 4.95 g of ethene (C₂H₄) are combusted with 3.25 g of oxygen.

a) What is the limiting reagent?

C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

• Only 0.102 moles of O₂ are available (need 0.531).

• O₂ is the limiting reagent.

b) How many grams of CO₂ are formed?

• From the balanced equation, 3 moles O₂ produce 2 moles CO₂.

• Moles of CO₂ = 0.102 × (2/3) = 0.068

• Mr of CO₂ = 44

• Mass = 0.068 × 44 = 2.99 g

3. Consider the reaction: C₆H₆ + Br₂ → C₆H₅Br + HBr

a) What is the theoretical yield of C₆H₅Br if 42.1 g of C₆H₆ react with 73.0 g of Br₂?

• Only 0.456 moles of Br₂ available.

• Br₂ is limiting.

Theoretical mass of C₆H₅Br:

• Mr of C₆H₅Br = 157

• Mass = 0.456 × 157 = 71.6 g

b) If the actual yield of C₆H₅Br is 63.6 g, what is the percent yield?

• Percentage yield = (63.6 ÷ 71.6) × 100 = 88.8%

4. Iron forms when iron(III) oxide is heated with carbon.
Fe₂O₃(s) + 3C(s) → 2Fe(s) + 3CO₂(g)

a) State how the equation shows that the iron(III) oxide is reduced.

• The Fe₂O₃ loses oxygen and forms Fe, meaning it undergoes reduction (loss of oxygen).

b) State why CO₂ must not be released to the atmosphere.

• CO₂ is a greenhouse gas and contributes to climate change and global warming.

c) Calculate the maximum mass, in tonnes, of iron that can be produced when 30 tonnes of iron(III) oxide are reacted with excess carbon.

• Moles of Fe₂O₃ = 30,000 ÷ 160 = 187.5

• Moles of Fe = 187.5 × 2 = 375

• Mass of Fe = 375 × 56 = 21,000 kg = 21 tonnes

d) A mixture of 25,000 moles of iron(III) oxide and 840,000 tonnes of carbon is heated. Which reactant is in excess?

• Moles of C needed = 25,000 × 3 = 75,000

• Available C:
  Mass = 840,000,000 kg
  Moles = 840,000,000 ÷ 12 = 70,000,000 moles

• Since 70,000,000 moles of C are available and only 75,000 needed,
  Carbon is in excess.

Hydrated Compounds

• Many salts exist in hydrated forms, meaning water is part of their crystal structure — not that they are wet or dissolved.
• This is shown in the formula as .xH₂O, e.g., CuSO₄·5H₂O.
• Salts can also exist in anhydrous forms, without water of crystallisation.
• You may recall anhydrous copper(II) sulfate and cobalt(II) chloride being used to test for water.

The number of water molecules can be found by a simple heating experiment:
CuSO₄·5H₂O (s) → CuSO₄ (s) + 5H₂O (g)

Method:

1. Weigh an empty crucible.

2. Add hydrated compound and reweigh.

3. Heat the crucible strongly.

4. Cool and reweigh.

5. Repeat heating and weighing until mass stays constant, ensuring all water is removed.

• Mass of water lost = 3.20 − 1.80 = 1.40 g

To find the number of water molecules:

• Therefore, the empirical formula = CuSO₄·7H₂O.

Determination of the Percentage and Moles of Water of Crystallisation in Hydrated Magnesium Sulphate

Safety

• Always wear eye protection.

Apparatus and Materials
• Balance (accurate to 2 decimal places)
• Spatula
• Hydrated magnesium sulphate
• Bunsen burner, tripod, gauze, heat resistant mat
• Evaporating basin
• Tongs

Procedure

1. Weigh a dry, clean evaporating basin.

2. Add about four spatulas of hydrated magnesium sulphate, then reweigh.

3. Record the masses.

4. Place the evaporating basin on the tripod and gauze.

5. Heat gently at first, then strongly after a few minutes.

6. Weigh the basin and its contents.

7. Reheat and reweigh until a constant mass is achieved.

8. Calculate the mass of water lost.

9. Determine the percentage of water of crystallisation using the data.
   (Ar: Mg = 24; S = 32.1; O = 16.0; H = 1)

Possible Errors

• Incomplete removal of water (not heated long enough).

• Loss of solid when heating strongly.

• Atmospheric moisture reabsorbed before weighing.

Results

Calculate the percentage of water of crystallisation:

Use mass and Mr to find the empirical formula:

• Empirical formula: MgSO₄·9H₂O

Why did we continue heating and reweighing until a constant mass?

• To ensure all the water of crystallisation has been removed and the reaction is complete.

• To ensure all water has been removed

• If not, the mass of anhydrous magnesium sulphate wool be larger than actual. This wold make the difference between the mass at the start and end smaller and so the mass of water loss would be smaller