# Question 106e3

May 3, 2015

You'd use that number like you would a whole number.

The stoiciometric coefficients of a balanced chemical equation tell you the proportion that exists for the species that take part in that reaction.

More specificially, they tell you the ratio that exists between the moles of each species that react - this is know as the mole ratio.

So, let's say you're dealing with the combustion of methane, $C {H}_{4}$. The balanced chemical equation is

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

Notice that you need 1 mole of methane to react with $\textcolor{red}{2}$ moles of oxygen in order to produce 1 mole of carbon dioxide and 2 moles of water.

In other words, regardless of how many moles of methane react, you'll always need twice as many moles of oxygen for the reaction to take place.

If you have 10 moles of methane, you'll need 20 of oxygen. If you have 0.25 moles of methane, you'll need 0.50 of oxygen. And so on.

Let's say you calculated the number of moles of oxygen to be 0.1374. Using the mole ratio that exists between the species, you'll get

0.1374cancel("moles "O_2) * ("1 mole "CH_4)/(color(red)(2)cancel("moles "O_2)) = "0.06870 moles" $C {H}_{4}$

0.1374cancel("moles "O_2) * ("1 mole "CO_2)/(color(red)(2)cancel("moles "O_2)) = "0.06870 moles" $C {O}_{2}$

0.1374cancel("moles "O_2) * ("2 moles "H_2O)/(color(red)(2)cancel("moles "O_2)) = "0.1374 moles "# ${H}_{2} O$

The mole ratios that exist between the species that take part in a chemical reaction are always true, regardless of how many moles of a reactant or product you start/end up with.