What is the maximum number of moles of H2O that can be produced from the reaction of 8.3 mol H2 and 2.9 mol O2?

H2 + O2 H2O

What is the limiting reactant?

What is the excess reactant?

The answer to your question is Moles of water formed = 5.8; limiting reactant = Oxygen, excess reactant = hydrogen

Explanation:

Data

moles of water = ?

moles of H₂ = 8.3 moles

moles of O₂ = 2.9 moles

Balanced chemical reaction

                  2H₂  +  O₂  ⇒  2H₂O

Process

1.- Calculate the limiting reactant

Theoretical yield =   2 moles of H₂ / 1 mol of O₂ = 2

Experimental yield = 8.3 moles / 2.9 moles = 2.9

As the proportion increased in the experimental yield, we conclude that the limiting reactant is Oxygen.

2.- Moles of H₂O form

               1 mol of O₂ 2 moles of water

               2.9 moles of O₂  x

               x = (2.9 x 2) / 1

              x = 5.8 moles of water

3.- excess reactant is H₂

O2 is the limiting reactant

H2 is the excess reactant

The maximum number of moles of H2O that can be produced is 5.8 moles

Explanation:

Step 1: Data given

Number of moles H2 = 8.3 moles

Number of moles O2 = 2.9 moles

Step 2: The balanced equation

2H2 + O2 → 2H2O

Step 3: Calculate the limiting reactant

For 2 moles H2 we need 1 mol O2 to produce 2 moles H2O

O2 is the limiting reactant. It will completely be consumed (2.9 moles). H2 is in excess. There will react 2*2.9 = 5.8 moles

There will remain 8.3 - 5.8 = 2.5 moles H2

Step 4: Calculate moles H2O

For 2 moles H2 we need 1 mol O2 to produce 2 moles H2O

For 2.9 moles O2 we'll have 2*2.9 = 5.8 moles H2O

The maximum number of moles of H2O that can be produced is 5.8 moles