Solving moles-to-moles limiting reactant problems

Question

hydrogen H2 

gas 

and oxygen O2 

gas react to form water H2O

 vapor. Suppose you have 3.0

 mol

 of H2

 and 11.0

 mol

 of O2

 in a reactor.Calculate the largest amount of 

H2O

 that could be produced. Round your answer to the nearest 0.1

 mol

.

Answer 1

First, write a correctly balanced equation for the reaction taking place.

2H₂(g) + O₂(g) ==> 2H₂O(g)

To find the limiting reactant, one easy way is to divide the mols of each reactant by the corresponding coefficient in the balanced equation:

3.0 mols H₂ divided by 2 = 1.5

11.0 mols O₂ divided by 1 = 11.0

Since 1.5 is less than 11.0, H₂ is limiting

The limiting reactant will determine the amount of product that can be produced. This will be according to the stoichiometry of the balanced equation.

3.0 mols H₂ x 2 mols H₂O / 2 mols H₂ = 3.0 mols H₂ produced