What is your limiting reactant if 32 grams of CH4 reacts with 32 grams of oxygen gas?

1 Answer
Apr 3, 2018

Oxygen is our limiting reactant.


We must start by creating a balanced reaction between Methane and Oxygen. As methane is a hydrocarbon reacting with oxygen this will be a combustion reaction resulting in carbon dioxide and water.

The combustion reaction is:

#CH_4 + 2O_2 ->CO_2 + 2H_2O#
Now we want to find how many moles of each reactant we have to find which one is limiting.

If we do a rough calculation for finding how many moles of #CH_4# and how many moles of #O_2# we have we get the following:
Using the molarity equation:

M=molar mass of the compound/element,
m=mass in grams of the compound/element
n=number of moles of the compound/element

n(Oxygen)=#32/32=1# mol

n(methane)=#(32/16)=2# mol

However, in the reaction equation, we need 2 moles of oxygen for every one mole of Methane, and we only have 1 mole of oxygen and 2 moles of Methane. That means methane is in excess and oxygen is thus our limiting reactant.

So in this reaction we would only be able to react 1 mol of #O_2# with 0.5 mol of #CH_4#.