CH_4 +2O_2 --> CO_2 + 2H_2O

Question

if you change 2O_2 to a 3O_2 what will be the change in the results on the right side of the eqaution

Answer 1

The chemical equation you provided is the combustion of methane (CH₄):

$${\text{CH}}_4+2{\text{O}}_2\to {\text{CO}}_2+2{\text{H}}_2\text{O}$$

This equation indicates that one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water.

If you change $2{\text{O}}_2$ to $3{\text{O}}_2$, the equation would be:

$${\text{CH}}_4+3{\text{O}}_2\to {\text{CO}}_2+\text{?}$$

To balance the equation with 3 moles of $O_2$, we need to recalculate the products. Each molecule of $O_2$ provides oxygen atoms for the products:

1. You start with 3 moles of $O_2$, which gives you 6 oxygen atoms.
2. The $C{O}_2$ product requires 2 oxygen atoms (1 carbon dioxide molecule).
3. That leaves you with $6-2=4$ oxygen atoms for water.
4. Each water molecule requires 1 oxygen atom, so you can form $4/2=2$ water molecules.

Thus, the balanced reaction with $3O_2$ becomes:

$${\text{CH}}_4+3{\text{O}}_2\to {\text{CO}}_2+2{\text{H}}_2\text{O}$$

However, since you've added more oxygen but didn't add more methane, the overall result in terms of products (the number of $C{O}_2$ and $H_{2}O$ molecules) doesn't actually change for this specific reaction under typical conditions. In the end, you still produce 1 molecule of $C{O}_2$ and 2 molecules of $H_{2}O$ when using $C{H}_4$. The excess $O_2$ would remain unreacted:

So the result is: 1 $C{O}_2$ and 2 $H_{2}O$, with 1 extra $O_2$ unreacted.