Reacting Masses, Gas Volumes & Limiting Reactants
Section: 3. Stoichiometry | Syllabus: Cambridge IGCSE Chemistry (0620)
Reacting Mass Calculations
A balanced equation gives the mole ratio in which substances react. Combining this ratio with Mr values lets us calculate the mass of any reactant or product in a reaction.
Mass → Moles → use mole ratio from equation → Moles → Mass
- Write the balanced equation
- Calculate Mr for the substance you're given and the substance you want to find
- Convert the given mass to moles (moles = mass ÷ Mr)
- Use the mole ratio from the equation to find moles of the unknown substance
- Convert those moles back to mass (mass = moles × Mr)
Worked Example: Basic Reacting Mass
- Question: what mass of magnesium oxide is produced when 12 g of magnesium burns completely in oxygen? 2Mg + O₂ → 2MgO (Ar: Mg = 24, O = 16) [3]
- Answer: moles Mg = 12 ÷ 24 = 0.5 mol [1]; ratio Mg:MgO is 2:2 (1:1), so moles MgO = 0.5 mol [1]; Mr of MgO = 40, so mass = 0.5 × 40 = 20 g [1]
When a question says a reactant is "in excess", it means there is more than enough of it, so every bit of the other (limiting) reactant reacts completely - you only need to calculate using the non-excess reactant.
Molar Gas Volume
At room temperature and pressure (r.t.p.), one mole of any gas occupies the same volume: 24 dm³ (24,000 cm³). This is the molar gas volume, and it applies to any gas, regardless of its identity.
- moles of gas = volume of gas (dm³) ÷ 24
- volume of gas (dm³) = moles of gas × 24
- Remember to convert cm³ to dm³ by dividing by 1000 before using these formulae
At r.t.p., 1 mole of any gas - collected in a syringe or measuring cylinder - occupies 24 dm³
Worked Example: Gas Volume from Moles
- Question: calculate the volume, at r.t.p., occupied by 0.2 mol of carbon dioxide gas. [1]
- Answer: volume = moles × 24 = 0.2 × 24 = 4.8 dm³ [1]
Worked Example: Reacting Masses Involving a Gas Volume
- Question: what volume of carbon dioxide, at r.t.p., is produced when 5 g of calcium carbonate decomposes completely? CaCO₃ → CaO + CO₂ (Ar: Ca = 40, C = 12, O = 16) [3]
- Answer: Mr of CaCO₃ = 100; moles CaCO₃ = 5 ÷ 100 = 0.05 mol [1]; ratio CaCO₃:CO₂ is 1:1, so moles CO₂ = 0.05 mol [1]; volume = 0.05 × 24 = 1.2 dm³ [1]
Limiting Reactants
When two reactants are mixed, the amounts present rarely match the mole ratio in the equation exactly. The reactant that runs out first is the limiting reactant - it limits how much product can form. The other reactant is in excess, and some of it is left over unreacted.
Method for Identifying the Limiting Reactant
- Calculate the moles of each reactant actually present
- Use the mole ratio from the equation to work out how much of one reactant would be needed to react completely with the other
- Compare this to how much is actually available - whichever reactant would run out first is limiting
- Use the moles of the limiting reactant (never the excess one) for all further calculations
Worked Example: Identifying the Limiting Reactant and Gas Volume Produced
- Question: 4.55 g of zinc is reacted with 50 cm³ of 2.25 mol/dm³ dilute hydrochloric acid. Zn + 2HCl → ZnCl₂ + H₂. Calculate the volume of hydrogen gas, at r.t.p., produced. (Ar: Zn = 65) [5]
- Answer: moles Zn = 4.55 ÷ 65 = 0.07 mol [1]; moles HCl = 2.25 × (50÷1000) = 0.1125 mol [1]; ratio Zn:HCl is 1:2, so 0.07 mol Zn would need 0.14 mol HCl to react completely - only 0.1125 mol HCl is available, so HCl is the limiting reactant [1]; moles H₂ = moles HCl ÷ 2 = 0.1125 ÷ 2 = 0.05625 mol [1]; volume H₂ = 0.05625 × 24 = 1.35 dm³ [1]
Common Mistakes
- ❌ Assuming the reactant given the smaller mass or volume is automatically the limiting reactant
- ✓ you must convert both reactants to moles and compare using the equation's mole ratio - a smaller mass is not automatically fewer moles, and the ratio itself may not be 1:1
- ❌ Using the moles of the excess reactant to calculate the amount of product
- ✓ product amount is always governed by the limiting reactant, since the excess reactant doesn't fully react
- ❌ Forgetting that 24 dm³ applies only to gases at r.t.p., not to solids, liquids, or solutions
- ✓ the molar gas volume relationship only converts between moles and volume for a substance in the gas state
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