By Riti Gupta
Updated Aug 30, 2022
In a chemical reaction, the reactant that is completely consumed first stops the reaction. That reactant is known as the limiting reactant (or limiting reagent). The remaining reactant, if any, is called the excess reactant.
In most laboratory settings, reactants are added in quantities that do not match the exact stoichiometric ratios shown in the balanced equation. Understanding which species is limiting helps chemists predict product yields, design scale‑up processes, and avoid costly waste.
The balanced equation for urea synthesis is:
\(2\mathrm{NH}_3(g)+\mathrm{CO}_2(g) \rightarrow (\mathrm{NH}_2)_2\mathrm{CO}(aq)+\mathrm{H}_2\mathrm{O}(l)\)
From the coefficients, 2 mol of ammonia react with 1 mol of carbon dioxide. Suppose we add 4 mol of \(\mathrm{CO}_2\) and 12 mol of \(\mathrm{NH}_3\).
Calculate the ammonia needed for 4 mol of carbon dioxide:
\(4\text{ mol }\mathrm{CO}_2\left(\dfrac{2\text{ mol }\mathrm{NH}_3}{1\text{ mol }\mathrm{CO}_2}\right)=8\text{ mol }\mathrm{NH}_3\)
Only 8 mol of ammonia are required, leaving 4 mol unused. Therefore, carbon dioxide is the limiting reactant and ammonia is in excess.
Alternatively, determine how much carbon dioxide would be needed for 12 mol of ammonia:
\(12\text{ mol }\mathrm{NH}_3\left(\dfrac{1\text{ mol }\mathrm{CO}_2}{2\text{ mol }\mathrm{NH}_3}\right)=6\text{ mol }\mathrm{CO}_2\)
Because only 4 mol of \(\mathrm{CO}_2\) are present, the same conclusion follows: carbon dioxide limits the reaction.
The balanced equation is:
\(2\mathrm{Al}+3\mathrm{Cl}_2 \rightarrow 2\mathrm{AlCl}_3\)
Given 25 g of aluminum and 32 g of chlorine gas, convert masses to moles:
Aluminum: \(25\text{ g }\mathrm{Al}\left(\dfrac{1\text{ mol }\mathrm{Al}}{26.98\text{ g }\mathrm{Al}}\right)=0.93\text{ mol }\mathrm{Al}\)
Chlorine: \(32\text{ g }\mathrm{Cl}_2\left(\dfrac{1\text{ mol }\mathrm{Cl}_2}{70.90\text{ g }\mathrm{Cl}_2}\right)=0.45\text{ mol }\mathrm{Cl}_2\)
Determine the chlorine required to consume all the aluminum:
\(0.93\text{ mol }\mathrm{Al}\left(\dfrac{3\text{ mol }\mathrm{Cl}_2}{2\text{ mol }\mathrm{Al}}\right)=1.40\text{ mol }\mathrm{Cl}_2\)
Only 0.45 mol of chlorine are available, so chlorine is the limiting reactant. The excess reactant is aluminum.
Verification: Moles of aluminum needed for all the chlorine:
\(0.45\text{ mol }\mathrm{Cl}_2\left(\dfrac{2\text{ mol }\mathrm{Al}}{3\text{ mol }\mathrm{Cl}_2}\right)=0.30\text{ mol }\mathrm{Al}\)
Since 0.93 mol of aluminum are present, the conclusion remains unchanged.
The limiting reactant is determined by the actual amounts of reactants used, not by the stoichiometric coefficients alone. Calculating the ratio of available moles to the stoichiometric ratio will always reveal which species limits product formation.
