Question

When this chemical equation is correctly balanced, what is the coefficient of the carbon dioxide molecule?

Answer 1

Surely it is `3`. `C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) +4H_2O(g)`

Explanation:

The balanced chemical equation for this combustion reaction is:

`C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) +4H_2O(g)`

Now this equation is stoichiometrically balanced: it is balanced with respect to charge as well as mass. For every carbon atom reactant particle, there is a carbon atom product particle; the same for hydrogen; and the same for oxygen.

Now you might not see this at the moment, but if you succeed in chemistry, and I think you will, you will become very adept at looking these sorts of reactions, and establishing balance, i.e. stoichiometry. Stoichiometry is a fancy word meaning `"in the right measure"`, and we practise stoichiometry all the time. When we make a cash transaction, we usually give a bank note to the vendor, the shopkeeper. The value of the goods, PLUS the value of the change, must equal to the value of the bank note. I suggest that when you buy stuff, you would know immediately whether you had received the right change. So, in fact, you already know about stoichiometry.

Likewise, in the banking caper, when you write a cheque (a debit item), there must be some corresponding credit item made to someone's account (i.e. your creditor). For every credit, there must be an equal and corresponding debit. Of course, sometimes debit and credit items don't add up. That's when you end up on a call queue to complain about a `£99-00` debit made to your account for a `£9-99` purchase.

So try this out. Represent the combustion of `"pentane",C_5H_12`, and `"hexane",C_6H_14` with oxygen gas. Carbon dioxide and water are of course the combustion products, but try balancing them correctly. One will be easy, the other may cause a problem. They are both doable on the basis of your example.

Check out here for another variation on this theme.