Question

How does stoichiometry operate in chemical reactions?

Answer 1

I hope to be of some help. Let's try the combustion of methane as an example of a chemical reaction.

Explanation:

The balanced equation for the combustion of methane is given by:

`underbrace(CH_4(g) +2O_2(g))_"80 g" rarr underbrace(CO_2(g) + 2H_2O(l))_"80 g"`

This equation is balanced stoichiometrically. What does it tell us? It tells us that `1` `mol` of methane gas and `2` `mol` of oxygen gas combine to form `1` `mol` of carbon dioxide, and `2` `mol` of water.

Because of the molar equivalence, i.e. the mass of a given mole of stuff, we could also say that `16*g` of methane combines with `64*g` oxygen gas, to give `44.0*g` of carbon dioxide, and `36` `g` of water.

It is worth noting that here that mass is conserved in every chemical equation. You start with `80` `g` of reactant (which we did!) you must finish with `80` `g` of product, which we did. The balanced chemical equation precisely specifies this molar and mass equivalence.

Such stoichiometry is also practised in other scenarios: banking, finance, accounting, even a simple transaction at the supermarket. You buy goods worth `£20-00` at the supermarket. You can present a `£20-00` note to the cashier, and the transaction is stoichiometrically balanced. Alternatively, you can present a card, and the `£20-00` is debited from your account, and credited to the supermarket's account - clearly a stoichiometric transaction, because credit item equal debit item.

This same stoichiometry operates in every chemical equation.

`"Garbage in EQUALS GARBAGE OUT"`

To end, you might say to me that you don't know how to balance a hydrocarbon combustion. I can guarantee that you know how to balance a cash transaction when you go shopping. You KNOW when you have been short-changed; likewise, you know when you have received too much change (and of course you give it back!). Don't short change your chemical equations; stoichiometry always operates.