Question

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

Answer 1

Oxygen is our limiting reactant.

Explanation:

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:

`n=(m)/M`
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`.