# STOICHIOMETRY CALCULATIONS

Consider the combustion of methane-

Thus, according to the above chemical reaction,

• One mole of CH4 (g) reacts with two moles of O2 (g) to give one mole of CO2 (g) and two moles of H2O(g)
• One molecule of CH4 (g) reacts with 2 molecules of O2 (g) to give one molecule of CO2 (g) and 2 molecules of H2O(g)
• 22.7 L of CH4 (g) reacts with 45.4 L of O2 (g) to give 22.7 L of CO2 (g) and 45.4 L of H2O(g)
• 16 g of CH4 (g) reacts with 2×32 g of O2 (g) to give 44 g of CO2 (g) and 2×18 g of H2O (g).

From these relationships, the given data can be interconverted as follows:

Limiting Reagent

• The reactant which is entirely consumed when a reaction goes to completion is called limiting reagent.
• It is present in smaller amount than calculated by balanced chemical equation.

Excess reagent

• The reactant which is not completely consumed in the reaction is called excess reactant. It is present in larger amount than calculated by balanced chemical equation.

For example, we calculate the mass of water formed when  of  reacts with  of  to form water in the following way:

H2 is limiting reagent because it is being entirely consumed.
Amount of Oleft unreacted
= 29 g – 24 g = 5 g

O2 is not consumed completely, therefore, it is excess reactant.

4 g of H2 will react with 32 g of O2 to form 36 g of H2O

3 g of H2 will react with 24 g of O2 to form