Reacting Masses and Limiting Reactants
Given the mass of one substance, work out the mass of another. The method never changes — and the one genuinely new idea is deciding which reactant runs out first.
1. The method
1. Write the BALANCED equation. 2. Convert the given mass to MOLES. 3. Use the MOLE RATIO from the equation. 4. Convert the moles back to MASS.
Check the equation is balanced first. Tutors flagged this as step zero — the whole calculation rests on the coefficients, so a wrong equation guarantees a wrong answer.
Worked example. What mass of magnesium oxide forms from 4.8 g of magnesium?
2Mg + O₂ → 2MgO
- Moles of Mg = 4.8 ÷ 24 = 0.2 mol
- Ratio Mg : MgO is 2 : 2 = 1 : 1, so moles of MgO = 0.2 mol
- Mr of MgO = 24 + 16 = 40
- Mass = 0.2 × 40 = 8 g
Build the ratio directly from the balanced equation’s coefficients. Confusion about how to build a ratio from an equation was recorded — the numbers in front of the formulae are the ratio.
Don’t try to go from mass to mass. A recorded error attempted exactly that. Masses do not scale with the coefficients; moles do.
2. Limiting reactants
When you are given the amounts of two or more reactants, one usually runs out first.
The LIMITING REACTANT is the one that is completely used up. It determines how much product can form. The reactant left over is IN EXCESS.
The product depends ONLY on the limiting reactant. Adding more of the excess reactant makes no difference — a recorded error missed that the same amount of zinc sulfide forms even when more sulfur is present.
How to identify it
1. Find the MOLES of each reactant. 2. DIVIDE each by its coefficient in the balanced equation. 3. The SMALLEST answer is the LIMITING reactant.
You must divide by the coefficient — comparing raw moles is not enough. Misunderstanding the role of stoichiometry in finding the limiting reactant was recorded. If a reaction needs 3 moles of one substance per mole of another, having equal moles of each means the first is limiting.
Worked example. 10 g of CaCO₃ reacts with 5 g of HCl. Which is limiting?
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
- Moles CaCO₃ = 10 ÷ 100 = 0.10; moles HCl = 5 ÷ 36.5 = 0.137
- Divide by coefficients: CaCO₃ → 0.10 ÷ 1 = 0.10; HCl → 0.137 ÷ 2 = 0.0685
- HCl is limiting; CaCO₃ is in excess
Note that HCl has more moles but is still limiting, because the equation needs twice as much of it. This is exactly why step 2 matters — and a recorded error assumed HCl was in excess for precisely this reason.
Shortcuts the question gives you
If the question says a reactant is “in excess”, it is telling you which one. The other is limiting. Tutors flagged this directly.
If only ONE reactant mass is given, that reactant is limiting — everything else is assumed to be in excess.
Always calculate the product from the LIMITING reactant. Using the excess reactant’s moles is the classic error, and it gives an answer that is too large.
3. Signs of excess in practice
How you can tell experimentally:
If a solid reactant remains undissolved at the end, it was in excess. If bubbling stops while solid remains, the other reactant has run out.
Continued effervescence does not indicate excess zinc — a recorded error. Effervescence means the reaction is still going; it is the leftover solid after bubbling stops that shows excess.
Explaining why something is in excess: state that there was more than enough moles of it to react with all of the other reactant, according to the mole ratio.
4. Worked example — full limiting reactant question
What mass of CO₂ forms when 10 g of CaCO₃ reacts with 5 g of HCl?
From above, HCl is limiting with 0.137 mol.
- Ratio HCl : CO₂ = 2 : 1
- Moles of CO₂ = 0.137 ÷ 2 = 0.0685 mol
- Mr of CO₂ = 44
- Mass = 0.0685 × 44 = 3.0 g (2 s.f.)
5. Percentage yield and percentage purity
% yield = (actual yield ÷ theoretical yield) × 100
The theoretical yield is what your reacting-mass calculation predicts; the actual yield is what was really obtained.
Yield is less than 100% because: the reaction may be reversible, there may be side reactions, or product is lost during transfer and purification.
% purity = (mass of pure substance ÷ total mass of sample) × 100
Percentage yield can never exceed 100%. If it does, check which yield you divided by.
6. Mistakes that cost marks
Not balancing the equation.
Converting mass to mass without the mole ratio.
Comparing raw moles without dividing by the coefficients.
Calculating the product from the excess reactant.
Missing the hint when a question says “in excess”.
Taking effervescence as evidence of excess.
Wrong Mr — miscounted atoms.
Rounding too early.
Missing units on the final mass.
Frequently asked questions
What is the limiting reactant? The one completely used up, which determines how much product forms.
How do I identify it? Find the moles of each, divide by its coefficient, and take the smallest.
Why divide by the coefficient? Because the equation may need more of one reactant than the other.
Which reactant do I use to calculate the product? Always the limiting one.
What if the question says a reactant is in excess? Then the other one is limiting.
What if only one mass is given? That reactant is limiting.
How can I tell experimentally that something was in excess? Solid remains after the reaction has stopped.
What is percentage yield? (actual ÷ theoretical) × 100.
Why is yield less than 100%? Reversible reactions, side reactions, and losses during transfer.
Can yield exceed 100%? No.
Quick revision checklist
- I balance the equation first
- I convert mass to moles before anything else
- I build the mole ratio from the coefficients
- I never convert mass to mass directly
- I can define limiting and excess reactants
- I divide moles by the coefficient to find the limiting reactant
- I calculate products from the limiting reactant only
- I spot the hints — “in excess”, or only one mass given
- I can explain excess from experimental observations
- I can calculate percentage yield and purity
- I know why yield is below 100%
- I give units and sensible significant figures
These notes cover reacting masses and limiting reactants in the Cambridge IGCSE Chemistry (0620) syllabus and are written for Grade 9–11 / Year 10–11 students. They are based on teaching patterns observed across a large set of one-to-one IGCSE Chemistry lessons, with particular attention to the errors students make most often and the wording examiners reward. Always check the current syllabus and data booklet for your own exam series.
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