Step by step on how to find limiting reactant, with an example !

In order to find a limiting reactant, you'll at least need the mass of both reactants, or even better, number of moles of it (but you can work this out with the mass).

1. Write a balanced equation for the reaction. Balancing the equation is especially important because the mole ratio (the numbers when balancing) can affect the answer.

2. Find the number of moles of each reactant. You can find this by dividing the mass in grams by the molar mass of substance (adding all atomic masses of all elements present).

3.  Find the actual number of moles needed in the reaction by considering the mole ratio from the equation. Which means, if the mole ratio of reactant A : reactant B is 1:1, you don't need to change anything, but, if its somewhat 2:1 or 2:3 etc, you have to multiply or divide the n.o.m.s to make the mole ratio balanced. (You can refer to example if this part is a bit confusing)

4.  Determine which reactant has the least number of moles. That one is the limiting reactant!

Example (Question took from 2013 IGCSE Chemistry Paper3);

"4.8g of calcium is added to 3.6 g of water. The following reaction occurs.

Ca + 2H2O → Ca(OH)2 + H2

Find the limiting reactant."

1. The balanced equation is already given, so you're good for this step!

2. Find the n.o.m.s for the 2 reactants.

no. of moles of Ca

= 4.8 / 40.1

= 0.1197 mol

no. of moles of H2O

= 3.6 / (1.0 x 2 + 16.0)

= 0.2 mol

3. Consider the equation again,

From equation, mole ratio of Ca:H2O = 1:2

So, no. of moles of moles of Ca = 0.1197 x 2

= 0.2394 mol

4. Since 0.2 mol is less than 0.2394mol, H2O is the limiting reactant.

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