Now suppose, we have 4 mol of H_{2} and 3 mol of O_{2}. When 4 mol of H_{2} and 3 mol of O_{2} are allowed to react, 4 mol of H_{2} will react with 2 mol of O_{2}. There would be 1 mol of O_{2} in excess because the given amount of 4 mol of H_{2} has now been completely consumed. Therefore, the reaction will come to stop despite 1 more mol of O_{2} is available. Thus the product formation of H_{2}O is limited by the amount of H_{2} present. Therefore, H_{2} is called limiting reagent.

**Note:** O_{2} is the reagent in excess in this case.

Pingback:Percentage composition of compound … msa

Pingback:Stoichiometry, Long Questions

Pingback:Stoichiometry, Short Questions, Chemistry 11 … msa