I'm going to do this in a step wise fashion

so we can talk about what I'm doing in each step.

In the first step I don't need the

coefficients and the reactions to stoichiometry at all.

I'm only going to create the grams of butane into moles of butane.

So I'll take my 13.5 grams of butane, and

divide it by the molar mass that was up here,

58.123 grams per

mole. My grams will cancel, and

that results in 0.23227 moles of butane. But

of course, not all of those digits are significant.

So I'm going to remind myself that there was only three significant digits.

If I put in a little

line under that two.

Alright, now I know how many moles of butane there are.

But I need to know how many moles of oxygen there are, so what

I can do is set up a proportion using the reactions to wikiometry, ratios.

So looking at the reaction I'm going to change color.

I see that two moles of butane reacts with 13 moles of oxygen.

Do you see that? So I can write a calculation like this.

I can say, two moles of butane reacts with 13 moles of oxygen.