Solving moles-to-moles limiting reactant problems

Question

Methane gas and oxygen gas react to form carbon dioxide gas and water vapor. Suppose you have 13.0 mol of CH4 and 5.0 mol of O2 in a reactor. Calculate the largest amount of CO2 that could be produced. Round your answer to the nearest 0.1 mol.

Answer 1

Step 1 - Write an unbalanced chemical reaction equation

Methane is CH₄, oxygen gas is O₂, carbon dioxide is CO₂ and water vapor is H₂O so:

CH₄ + O₂ → CO₂ + H₂O

Step 2 - Balance the equation

CH₄ + 2O₂ → CO₂ + 2H₂O

Step 3 - Find the limiting reactant

The balanced chemical reaction equation tells you that it takes 1 mole of CH₄ for every 2 moles of O₂. So for 5 moles of O₂, it takes only 2.5 moles of CH₄ so there extra CH₄ available. That makes the O2 the limiting reactant.

Step 4 - Calculate the moles of CO₂ produced

The balanced chemical reaction equation tells you that for every 2 moles of O₂, there is 1 mole of CO₂ produced so for 5 moles of O₂, there are 2.5 moles of CO₂ produced (at the most).

Answer 2

First things first. We must write a correctly balanced equation:

Methane gas = CH₄(g)

Oxygen gas = O₂(g)

Carbon dioxide gas = CO₂(g)

Water vapor = H₂O(g)

Balanced equation is...

CH₄(g) + 2O₂(g) ==> CO₂(g) + 2H₂O(g)

Now, to find the limiting reactant, there are several ways to do this. Here I show one way that I find the easiest.

You divide the moles of each reactant by the corresponding coefficient in the balanced equation. The value that is the least represents the limiting reactant.

For CH₄(g) we have: 13.0 mols (÷1->13)

For O₂(g) we have: 5.0 mols (÷2->2.5) .. Therefore O₂ is the limiting reactant (2.5 < 13)

Next, we use the MOLES of the limiting reactant and the stoichiometry of the equation to find the amount of product (CO₂) that can be produced:

5.0 moles O₂ x 1 mole CO₂ / 2 mols O₂ = 2.5 mols CO₂