Butane gas is compressed and used as a liquid fuel in...

Butane gas is compressed and used as a liquid fuel in disposable cigarette lighters and lightweight camping stoves. Suppose a lighter contains $5.50 \mathrm{~mL}$ of butane $(d=0.579 \mathrm{~g} / \mathrm{mL})$ (a) How many grams of oxygen are needed to burn the butane completely? (b) How many moles of $\mathrm{H}_{2} \mathrm{O}$ form when all the butane burns? (c) How many total molecules of gas form when the butane burns completely?

Transcript

00:01 For this question, we have the combustion of butane, and we want to know first how many grams of oxygen is required to completely combust 5 .50 milliliters of butane.

00:10 Then we want to know from the same amount of butane how many moles of water are formed.

00:15 And then finally, what the total number of gas molecule products are that is formed from the combustion of this amount of butane.

00:22 So we're going to have four dimensional analysis calculations, and then we'll have one just addition calculation.

00:30 The first thing that we need to do is balance the chemical reaction now for this combustion of methane, or not methane, butane, because we're going to need the mole ratios that come from the balanced chemical reaction.

00:43 So to balance this, we're going to put a two in front of butane, 13 in front of oxygen, 10 in front of water, and then eight in front of carbon dioxide.

00:56 So now we can move on to our calculations.

00:58 So the first one, we're going to start with our 5 .5 milliliters of butane.

01:04 And then we're going to convert to grams of oxygen.

01:07 So the first thing we're going to use is butane's density to get to grams.

01:10 So we'll have 0 .579 grams over 1 milliliter.

01:17 Then we're going to use butane's molar mass to get to moles.

01:20 This will allow us to convert between substances.

01:22 So we're going to have one mole of butane on top.

01:27 And then we'll have a smaller mass on the bottom, which is 58.

01:31 1 .14 grams.

01:36 Then we're going to use a mole ratio so we can get two moles of oxygen.

01:40 So we're going to have 13 moles of oxygen on top, and then two moles of butane on the bottom.

01:47 And these numbers just come from our balance to chemical reaction.

01:50 And then finally, we're going to use oxygen's molar mass to get to grams.

01:53 So we'll have 32 grams of oxygen on top, and one mole on the bottom.

01:58 So we're going to multiply everything across the top and then divide by the things on the bottom.

02:02 And we're going to get 11 .4 of oxygen is needed.

02:07 So that's our first answer.

02:12 Next, we're going to calculate the number of moles of water that's produced when 5 .5 milliliters in butane is combusted.

02:20 So we'll start with our 5 .5 milliliters again.

02:26 And then we're going to use the density just like we did in the last calculation, as well as butane's molar mass, just like before.

02:42 And then for our next step, we're going to use a different mole ratio because we're trying to get to water...

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