moles-to-moles limiting reactant

Question

Methane (CH4) gas and oxygen (O2) gas react to form carbon dioxide (CO2) gas and water (H2O) vapor. Suppose you have 7.0 mol of CH4 and 1.0 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. Balance the reaction.

CH4 + O2 ~> CO2 + H2O

I see 4 H on the left side and 2 H on the right, so will add 2 in front of H2O.

CH4 + O2 ~> CO2 + 2H2O

Then I notice that I have a total of 4 O atoms on the right, but only two (O2) on the left, so will put a 2 in front of O2.

CH4 + 2O2 ~> CO2 + 2H2O

Now the reaction is balanced.

Step 2. Identify the limiting reagent.

The problem states that we have 7 moles of CH4 and 1 mol of O2. From our reaction coefficients, we can see a 1:2 stoichiometry for CH4:O2, meaning that for 7 moles of CH4, we would need 14 moles of O2 to fully react. Since we only have 1 mole of O2, O2 is the limiting reagent.

Step 3. Using the reaction coefficients and limiting reagent, calculate the total amount of product that can possibly be made.

Our stoichiometry is 1CH4:2O2:1CO2:2H2O

If we are limited by 1 mole of O2, we can calculate the amount of CH4 to be consumed, and products to be made, using simple ratios.

ex. 1CH4:2O2//XCH4:1O2

X = 1 x 1 / 2 = 0.5

Similarly, repeating this process for CO2:

ex. 2O2:1CO2//1O2:XCO2

X = 1 x 1 / 2 = 0.5

The total amount of CO2 that can be produced is 0.5 moles. Note that the question includes a line about rounding to the nearest 0.1 mol - that is intended to throw you off from the fact that the answer can be found via stoichiometry, and no complicated math is required.