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

Apr 3, 2018

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:

$C {H}_{4} + 2 {O}_{2} \to C {O}_{2} + 2 {H}_{2} O$
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 $C {H}_{4}$ and how many moles of ${O}_{2}$ we have we get the following:
Using the molarity equation:

$n = \frac{m}{M}$
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)=$\frac{32}{32} = 1$ mol

n(methane)=$\left(\frac{32}{16}\right) = 2$ 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 $C {H}_{4}$.