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A question gives you amounts of two reactants and asks how much product forms. Most mistakes happen before any real chemistry: comparing grams, or comparing moles without looking at the equation. This post gives a one-line test for the limiting reagent, then shows how to find the leftover reactant and the percentage yield.
Where is this in the syllabus?
The CBSE Chemistry curriculum for 2026-27 puts "Stoichiometry and Stoichiometric Calculations" in Unit 1 of Class 11, and the JEE (Main) 2026 syllabus lists "Chemical equations and stoichiometry". The NCERT Class 11 unit on basic concepts defines the idea in section 1.10.1: "the reactant, which gets consumed first, limits the amount of product formed and is, therefore, called the limiting reagent."
NCERT does not use the term percentage yield, so treat that part of this post as a short extension. If you want the five most common stoichiometry slips first, read our post on mole concept errors; this one goes further into leftovers and yield.
Molar masses below are rounded from the CIAAW standard atomic weights, for example 65.38 for zinc becomes 65.4.
How do you find the limiting reagent in one line?
Convert each reactant to moles, then divide by its coefficient in the balanced equation. The smallest number is the limiting reagent.
Worked example 1. 6.54 g of zinc is added to 150 mL of 1.0 M hydrochloric acid. Zn + 2HCl → ZnCl₂ + H₂.
Moles: Zn = 6.54 ÷ 65.4 = 0.100 mol. HCl = 0.150 L × 1.0 mol/L = 0.150 mol.
Divide by coefficients: Zn 0.100 ÷ 1 = 0.100; HCl 0.150 ÷ 2 = 0.075.
HCl gives the smaller number, so HCl is limiting.
H₂ formed = 0.150 ÷ 2 = 0.075 mol = 0.075 × 2.0 = 0.15 g.
Zinc used = 0.075 mol, so zinc left = 0.100 − 0.075 = 0.025 mol = 0.025 × 65.4 ≈ 1.64 g.
Notice the trap. Zinc has fewer moles (0.100 against 0.150), and a quick look would call it limiting. It is not, because each zinc needs two HCl.
How do you calculate percentage yield?
Real reactions rarely give the full amount. The theoretical yield is what the limiting reagent could make; the actual yield is what you collected.
Percentage yield = (actual yield ÷ theoretical yield) × 100
Worked example 2. 16.0 g of Fe₂O₃ reacts with 7.0 g of CO: Fe₂O₃ + 3CO → 2Fe + 3CO₂. Suppose 8.40 g of iron is collected (a sample number). Take Fe₂O₃ = 160, CO = 28, Fe = 56 and CO₂ = 44 g/mol.
Moles: Fe₂O₃ = 16.0 ÷ 160 = 0.100 mol. CO = 7.0 ÷ 28 = 0.250 mol.
Divide by coefficients: 0.100 ÷ 1 = 0.100; 0.250 ÷ 3 = 0.0833. CO is limiting.
Theoretical Fe = 0.250 × (2/3) = 0.1667 mol = 0.1667 × 56 ≈ 9.33 g.
Percentage yield = 8.40 ÷ 9.33 × 100 ≈ 90.0%.
Fe₂O₃ used = 0.250 ÷ 3 = 0.0833 mol ≈ 13.3 g, so about 2.7 g is left over.
Mass check. Reactants: 16.0 + 7.0 = 23.0 g. At full conversion the products are 9.33 g of iron and 0.250 × 44 = 11.0 g of CO₂, plus 2.67 g of unused Fe₂O₃: 9.33 + 11.0 + 2.67 = 23.0 g. The books balance.
How do you work backwards from a yield?
Worked example 3. How much CaCO₃ must you heat to collect 5.6 g of CaO if the yield is 80%? CaCO₃ → CaO + CO₂, with CaCO₃ = 100 and CaO = 56 g/mol.
Theoretical CaO needed = 5.6 ÷ 0.80 = 7.0 g = 7.0 ÷ 56 = 0.125 mol.
The equation is 1 : 1, so CaCO₃ = 0.125 mol = 12.5 g.
The common slip is to multiply by 0.80 instead of dividing. Ask yourself: with a poor yield, do I need more starting material or less? More, so divide.
Which mistakes cost marks?
Comparing grams. Grams of different substances cannot be compared; moles can.
Comparing moles without the coefficients. Example 1 shows how this flips the answer.
Using the excess reagent for the theoretical yield. Only the limiting reagent sets it.
Accepting a yield above 100%. A real yield cannot exceed the theoretical one, so check your arithmetic or the question.
Rounding too early. Keep 0.1667, not 0.17, until the last line.
A practice routine
Take five problems with two reactant amounts. For each, write the "divide by coefficient" line before anything else.
For every problem, also find the leftover reactant and do the mass check.
Do two yield problems forwards and one backwards, like example 3.
Once a week, mix these with empirical and molecular formula problems so you have to spot which method each question needs.
If you get a different answer from the key, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can find the line where your working went off.
Tonight, redo worked example 1 with 9.81 g of zinc and see whether the limiting reagent changes.
