What volume of air, measured at 298 K and 101 kPa, is...

What volume of air, measured at $298 \mathrm{K}$ and $101 \mathrm{kPa}$, is required to burn $2.00 \mathrm{kg} \mathrm{C}_{8} \mathrm{H}_{18}$ ? Air is approximately $78.1 \% \mathrm{N}_{2}$ and $20.9 \% \mathrm{O}_{2},$ by volume. Other gases make up the remaining $1.0 \%$.

Key Concepts

Ideal Gas Law

The ideal gas law (PV = nRT) is pivotal in relating the pressure, volume, temperature, and number of moles of a gas. It allows for the conversion from moles of oxygen (derived through stoichiometry) to the volume of air required under specified conditions of pressure and temperature.

Combustion Stoichiometry

This concept involves using a balanced chemical equation to relate the amounts of reactants and products in a chemical reaction. In the context of fuel burning, stoichiometry is used to determine how many moles of oxygen are required to completely oxidize a given amount of a hydrocarbon, thus linking the fuel’s mass with the necessary oxygen quantity.

Mole Concept

The mole concept is a fundamental principle in chemistry that provides a bridge between the atomic/molecular scale and macroscopic measurements. It allows one to convert quantities given in mass (for example, of a fuel) into moles, and thereby use stoichiometric ratios established by balanced chemical equations.

Composition of Air

Understanding the composition of air is crucial when calculating the actual volume of air needed since oxygen is only a fraction of air. Knowing the percentage of oxygen in air, along with the presence of other gases, helps accurately determine how much air must be ingested to obtain a sufficient amount of oxygen for the combustion process.

Transcript

00:17 In this question, we're given that we are required to burn 2 kilograms of c8h18 in excess oxygen.

00:29 We're asked what volume of air measured at 298 kelvin and 101 kilopascals is required to burn this amount.

00:42 So the first thing we're going to want to do is we're going to want to write a combustion reaction.

00:47 So we know with any combustion reaction, the products are always going to be co2 and h2.

00:55 Now first things first, we're going to need to balance this reaction.

01:01 So i see eight carbons on the left side, but only one on the right side.

01:05 So we're going to put an eight in front of this co2.

01:09 I also see 18 hydrogens on the right side, or i'm sorry, on the left side, but only two on the right side.

01:20 So we need to put a nine in front of here.

01:24 Now doing this gets us 16 plus 9, 25, oxygen on the right side, but only two on the left side.

01:36 So i'm going to go ahead and put a 12 .5 here.

01:41 Now doing this doesn't give us whole number moles.

01:43 So we're going to take this entire thing and multiply it by two.

01:57 So now we have our balanced equation.

02:04 We're given that we have two kilograms of c8h18.

02:14 So the first thing we're going to want to do is convert this into moles.

02:18 So two kilograms is 2 ,000 grams.

02:22 And we're going to divide it by the molar mass, which in this case we know is going to be 8 times 12 plus 1 times 18, which is equal to 114 grams...

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