# What is Limiting Reagents ?

## About Limiting reagent

In many situations one of the reactants is present in excess therefore some of this reactant is left over on completion of the reaction. For example, consider the combustion of hydrogen.

2H2(g) + O2(g) ⎯⎯→ 2H2O(g)

Suppose that 2 moles of H2 and 2 moles of O2 are available for reaction. It follows from the equation that only 1 mole of O2 is required for complete combustion of 2 moles of H2 ; 1 mole of O2 will, therefore, be left over on completion of the reaction. The amount of the product obtained is determined by the amount of the reactant that is completely consumed in the reaction. This reactant is called the limiting reagent. Thus, limiting reagent may be defined as the reactant which is completely consumed during the reaction.

In the above example H2 is the limiting reagent. The amount of H2O formed will, therefore, be determined by the amount of H2. Since 2 moles of H2 are taken, it will form 2 moles of H2O on combustion.

The best method to identify limiting reagent is by dividing given moles of each reactant by their stoichiometric coefficient, the one with least ratio is limiting reagent. It is particularly useful when number of reactants are more than two.

### Solved example of limiting reagent

How much magnesium sulphide can be obtained from 2.00 g of magnesium and 2.00 g of sulphur by  the reaction Mg + S→ MgS? Which is the limiting reagent? Calculate the amount of the reactants which remains unreacted.

Solution :  First of all each of this masses are converted into moles:

2.00 g of  Mg = = 0.0824 moles of Mg

2.00 g of S = = 0.0624 moles of S

From the equation, Mg + S → MgS, it follows that one mole of Mg reacts with one mole of S. We are given more moles of Mg than of S. Therefore, Mg is in excess and some of it will remain unreacted when the reaction is over. S is the limiting reagent and will control the amount of product. From the equation we note that one mole of S gives one mole of MgS, so 0.0624 mole of S will react with 0.0624 mole of Mg to form 0.0624 mole of MgS.

Molar mass of MgS         =    56.4 g

∴ Mass of MgS formed    =    0.0624 × 56.4 g = 3.52 g of MgS

Moles of Mg left unreacted     =    0.0824 –0.0624 moles of Mg

=     0.0200 moles of Mg

Mass of Mg left unreacted     =    moles of Mg × molar mass of Mg

=     0.0200 × 24.3 g of Mg = 0.486 g of Mg