Butane gas is compressed and used as a liquid fuel in dis- posable cigarette lighters and lightw

step 1 So we have a butane going to undergo a combustion reaction. We have oxygen to produce water and

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

Key Concepts

Reaction Stoichiometry

Reaction stoichiometry involves using the balanced chemical equation to determine the quantitative relationships between reactants and products. In combustion reactions, such as that of butane burning with oxygen, stoichiometry allows us to calculate how much oxygen is required based on the amount of fuel, and in turn determine the amounts of products like carbon dioxide and water formed.

Density and Volume Conversion

This concept is used to convert between volume and mass, based on the density of a substance. In problems where the volume of a liquid fuel is given, the density (mass per unit volume) can be used to calculate the mass of the substance, which is essential for subsequent stoichiometric calculations.

Grams to Moles Conversion

The conversion from grams to moles is vital in chemical calculations because stoichiometry is based on mole relationships as given by the balanced equation. Using the molar mass of a substance, the mass measured can be converted to moles, allowing the use of mole ratios to predict the amounts of other substances produced or consumed.

Avogadro's Number and Molecule Calculations

Avogadro's number provides a link between moles and the number of molecules. After finding the number of moles of a substance from stoichiometric calculations, this number can be multiplied by Avogadro's number to determine the total number of molecules involved in the reaction, which is especially useful in converting chemical amounts into discrete particle counts.

Transcript

00:01 So we have a butane going to undergo a combustion reaction.

00:04 We have oxygen to produce water and also co2.

00:08 And what would be the mass of oxygen required to completely burn the butane? and then for b, what would be the number of most of water form when all the butane burns? and also how many numbers of most of total gas being formed when all the butanings being burned.

00:26 Okay, so first of all we have to write the balanced chemical reactions.

00:30 We know that the butane is going to we have with oxygen to form water and co2 so we're going to balance it so first of all we balance carbon so i have four carbon so i have four carbon even of co2 and then we have 10 oxygen so we put up five in front of water to have 10 oxygen and thus but now is we are going to balance the oxygen so on the right we have five plus uh so we have five oxygen and then we have eight oxygen so we have 13 13 oxygen, 13 oxygen.

01:02 So this is an odd number.

01:04 So we are going to use a fashion here, just make things a little bit easier.

01:07 So we have 13 double by 2, so we have also have 13 oxygen on the left and on the right.

01:14 Okay, so the first question is, what would be the oxygen required to completely we have with all of the butane? assume we have 5 .5 milliter of butane, and also the density is 0 .579, 0 .579.

01:30 Okay, so the mass of building, so c4h10 mass, or which is m, equals to our, of course, equals to our volume times our density, so we have the mass of 5 .50, multiplied by 0 .579 gram per millimeter, and then we should know it's 5 .5 times 0 .579.

01:56 So it will be 3 .18 gram.

01:59 All right, so the next step, again, straightforward, we are going to convert all the mass weight.

02:04 We have two lumber most.

02:06 So we have 3 .18 divided by the molar mass of bultane.

02:12 So we have a 4 carbon and also with 10 hydrogen.

02:16 So it will be equals to 0 .0548 mole.

02:22 This would be the number of most of our bultane.

02:29 Okay, so we want to know the mass of our oxygen.

02:34 So actually, we are looking at the mass.

02:36 So if we want to know that, we have to find out the number of modes of oxygen, and then we can convert to a number of mass, you know, the food we have with our methane.

02:43 So our methane is our 0 .05478 modes, and the motor ratio of methane to carbon, not carbon outside, to your oxygen.

02:53 So we were looking for oxygen, so this oxygen is 6 .1 to 13 divided by 2.

02:59 So we just need to take 30 divided by 2 is 6 .5, so the ratio essentially equals to 1 to 6 .5.

03:05 So the number of move of our oxygen will be equals to our number of modes of our mutine times 6 .5.

03:17 And then we will have 6 .356 mole, and then we can convert that to the mass of oxygen.

03:27 You will just need to multiply that by the molar mass of oxygen.

03:32 So it will be 11 .4.

03:38 11 .4.

03:41 All right, so we solve part a.

03:44 Part b, what would be the number of move of water being formed when all the butane burns? again, we go back to our chemical reaction.

03:52 We find that the molar ratio between our methane and the water is 1 to 5, 1 to 5...

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