Divide the moles of each reactant by its coefficient in the balanced equation; the reactant giving the smallest value is the limiting reactant and decides how much product forms.
This page walks through limiting reactant for SPM Chemistry, step by step: the formula you need, the units to watch, and a worked example.
When you use this
When two amounts of reactant are given, one usually runs out first and stops the reaction. That is the limiting reactant, and every amount of product must be worked out from it, not from the reactant left over in excess.
The method and units
There is no single formula; there is a reliable method:
- Find the moles of each reactant (n = mass / molar mass, or n = MV, or n = V / molar volume).
- Divide each reactant’s moles by its coefficient in the balanced equation.
- The reactant with the smallest value is the limiting reactant.
- Use the moles of the limiting reactant, and the mole ratio, to find any product.
Units. Amounts in mol; the comparison in step 2 is a pure number. Only the limiting reactant enters the product calculation.
Worked example 1 (easy)
2H2 + O2 → 2H2O. You have 4 mol of hydrogen and 3 mol of oxygen. Which is limiting?
- H2: 4 / 2 = 2. O2: 3 / 1 = 3.
- The smaller value is for hydrogen, so hydrogen is limiting and oxygen is in excess.
Worked example 2 (medium)
Mg + 2HCl → MgCl2 + H2. You have 0.10 mol of magnesium and 0.15 mol of hydrochloric acid. Find the moles of hydrogen produced.
- Mg: 0.10 / 1 = 0.10. HCl: 0.15 / 2 = 0.075.
- HCl gives the smaller value, so hydrochloric acid is limiting.
- Ratio HCl : H2 = 2 : 1, so n(H2) = 0.15 / 2 = 0.075 mol.
Worked example 3 (SPM level)
Mg + 2HCl → MgCl2 + H2. 2.4 g of magnesium is added to 50 cm3 of 2.0 mol dm−3 hydrochloric acid. Find the volume of hydrogen at room conditions. (Mg = 24)
- n(Mg) = 2.4 / 24 = 0.10 mol.
- n(HCl) = 2.0 x (50 / 1000) = 0.10 mol.
- Mg: 0.10 / 1 = 0.10. HCl: 0.10 / 2 = 0.05. HCl is limiting.
- Ratio HCl : H2 = 2 : 1, so n(H2) = 0.10 / 2 = 0.05 mol.
- V(H2) = 0.05 x 24 = 1.2 dm3.
Why the excess reactant is a trap
Notice that in the last example equal moles of the two reactants did not mean they were used up together: because the equation needs twice as much acid as metal, the acid ran out while some magnesium remained. This is exactly why you must divide by the coefficient rather than compare raw moles, and why the leftover reactant plays no part in the amount of product formed.
Where it appears in the exam
Limiting reactant questions reward a clear layout. Show the moles of each reactant, show each one divided by its coefficient, state which is limiting in words, and then carry only that amount forward. Because the working makes your reasoning visible, method marks are awarded even if the final volume or mass is slightly out, so never jump straight to the answer. The same layout also helps when a question asks for the mass of the excess reactant left over, because you already know how much started and how much reacted.
Common traps
- Comparing raw moles. Always divide by the coefficient first; 0.10 mol Mg and 0.10 mol HCl are not a matched pair.
- Assuming the smaller mass is limiting. A small mass of a light substance can still be many moles.
- Using the excess reactant for the product. Product always comes from the limiting reactant.
- Stopping too early. After finding the limiting reactant, still apply the mole ratio to reach the answer the question wants.
Our teachers have students lay the two “moles divided by coefficient” figures side by side, so the limiting reactant is obvious at a glance and the rest of the calculation follows the standard mole-ratio route.
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