How does stoichiometry operate in chemical reactions?

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.