Butane gas is compressed and used as a liquid fuel in disposable cigarette lighters and lightwei

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

Key Concepts

Stoichiometry

Stoichiometry involves using a balanced chemical equation to determine the quantitative relationships between reactants and products in a reaction. For the combustion of a fuel like butane, it allows one to calculate the amount of oxygen required and the quantities of products formed, based on the mole ratios established by the balanced reaction.

Density and Volume-Mass Conversion

When a substance's volume and density are given, the mass can be calculated by multiplying the two values. This mass can then be used, along with the substance’s molar mass, to determine the number of moles present, which is a crucial step in stoichiometric calculations.

Mole Concept

The mole is a fundamental unit that links the mass of a substance to the number of entities it contains. In chemical calculations, converting mass to moles allows one to work directly with the coefficients from the balanced chemical equation, facilitating the determination of the amounts of reactants and products.

Avogadro's Number

Avogadro's Number, which is approximately 6.022 x 10^23, is used to convert the amount of substance in moles to the actual number of molecules or atoms. This conversion is essential when determining the total number of gas molecules produced in a reaction.

Transcript

00:01 For this problem, we're going to be using a balanced chemical equation to predict the outcome of a reaction.

00:05 We're given that we have 5 .50 milliliters of butane, which the chemical formula is c4h10, and we're given that the density of butane is 0 .79 grams per milliliter.

00:17 We also know that this is a combustion reaction.

00:20 So we're going to have to find how many grams of o2 butane needs to react completely, how many moles of water vapor are going to be produced, and then the total number of gaseous particles that are going to be produced in this reaction.

00:33 And for a combustion reaction, we know that it's going to be our fuel reacting with oxygen gas to produce co2 and water vapor.

00:41 So our first step is going to be to balance this equation.

00:45 And so we're going to write out c, h, and o, and then add up how many.

00:52 Of these on each side we have.

00:54 In the products we have one carbon, two hydrogen, and three oxygen, which is standard for a combustion reaction.

01:00 In the reactants we have four carbon, 10 hydrogen, and two oxygen.

01:07 I am going to balance our carbon first.

01:12 For these, we'll then produce, okay, so then we have nine oxygen.

01:16 We have two times four, which is eight plus one.

01:19 Over here is nine.

01:21 And then for our hydrogen, well, okay, and then this is going to be, this is going to become 13 because we have five plus the eight from the co2, which doesn't work.

01:34 So i'll put a two in front of the c4h10, which will then give us eight carbons, 20 hydrogens.

01:41 The oxygen still say the same.

01:43 I'm going to go ahead and update, i'm going to update our hydrogen first actually and put a 10 here which will give us our 20 and then that will be that's going to give us 18 oxygen but we're going to go ahead and update our co2 as well and make that 8 instead which will give us 8 yeah 8 carbon 16 oxygen plus 10 which would be 26 which then if we put 13 in front of the o2 we get our 26 so so this is our balanced equation that we're going to be using for everything else that we need to do.

02:22 So the first thing that we need to do is convert our milliliters into grams by using our density that we're given here.

02:30 So we start with our 5 .50 milliliters of butane.

02:39 And we convert it to grams by using the conversion factor.

02:44 We know that one millimeter is the same thing as 0 .579 grams per milliliter.

02:52 All right.

02:54 So we multiply 5 .50 times 0 .579, and that's going to give us 3 .18 grams of c4h10.

03:06 But we need to have this in moles because otherwise we can't do anything with it with our molar ratio.

03:12 So we need to find the molar mass of c4h10.

03:17 We have four carbon, 10, 10 hydrogen.

03:20 We multiply by the respective atomic masses.

03:24 Carbons is 12 .011, hydrogens is 1 .008.

03:30 When we multiply those out, we get 48 .044.