# How would you write a balanced chemical equation for CH4 reacting with oxygen gas to produce water and carbon dioxide?

Nov 13, 2015

$C {H}_{4}$ + $2$ ${O}_{2} \to$ $2$ ${H}_{2} O + C {O}_{2}$

$C {H}_{4}$ + $2 {O}_{2}$ $\rightarrow$ $C {O}_{2}$ + $2 {H}_{2} O$

This is the balanced reaction equation for the combustion of methane.

#### Explanation:

The Law of Conservation of Mass basically states that matter can neither be created nor destroyed. As such, we must be able to show this in our chemical reaction equations.

If you look at the equation above, you'll see an arrow that separates the reaction equation into two parts. This represents the direction of the reaction.

• To the left of the arrow, we have our reactants.

• To the right of the arrow, we have our products.

The quantity of each individual element in the left must equal the quantity of each individual element in the right.

So if you look below, you'll see the unbalanced equation, and I'll try to explain how to balance the reaction.

$C {H}_{4}$ + ${O}_{2}$ $\rightarrow$ $C {O}_{2}$ + ${H}_{2} O$

Our reactants in this equation are $C {H}_{4}$ and ${O}_{2}$.

Our next step is to break these down into individual atoms.

We have:

• 1 C atom, 4 H atoms & 2 O atoms.

If you're confused by this, look to see the little number to the bottom right of each element, the subscript, and it tells you how many of each atom are in the molecule. Make sense?

Now we look to the other side of the equation.

Here we see our products are $C {O}_{2}$ + ${H}_{2} O$

Again, we break these down into individual atoms again.

We have:

• 1 C atom, 2 H atom, 3 O atoms

This isn't right? What's wrong?

Yes, do you see it? We have more H atoms in the reactant side than the product side, and more O atoms in the product side than the reactant side. According to the Law of Conservation of Mass though, this isn't possible.

So how do you suppose we fix this?

We have to make the number of atoms on both sides equal, don't we? When we achieve this, we'll have a balanced equation.

To change the number of atoms, we can place a number in front of it, known as the coefficient. This multiplies the number of every atom by whatever number you use as the coefficient. BEWARE however, that we can never change the subscript (the little number), because doing so changes the chemical.

We have 1 C on both sides of the reaction - so this doesn't need to change.

We have 4 H on the left, and 2 H on the right. To make them equal, we place a 2 in front of the ${H}_{2} O$ like so; $2 {H}_{2} O$. This gives us 4 H on both sides, but it also gives us another O, making the total 4 O on the right....but we still only have 2O on the left, don't we?

Place a 2 in front of the ${O}_{2}$ to make 4 O on the left.

Now your reaction equation is balanced, and should look like this...

$C {H}_{4}$ + $2 {O}_{2}$ $\rightarrow$ $C {O}_{2}$ + $2 {H}_{2} O$