The hydrocarbon octane (C8 H18) burns to give CO2 and water vapor: 2 C8 H18(g)+25 O2(g) ⟶16 CO2(

The hydrocarbon octane (C8 H18) burns to give CO2 and water...

The hydrocarbon octane $\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)$ burns to give $\mathrm{CO}_{2}$ and water vapor: $$ 2 \mathrm{C}_{8} \mathrm{H}_{18}(\mathrm{g})+25 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 16 \mathrm{CO}_{2}(\mathrm{g})+18 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ If a $0.095-\mathrm{g}$ sample of octane burns completely in $\mathrm{O}_{2},$ what will be the pressure of water vapor in a $4.75-\mathrm{L}$. Thask at $30.0^{\circ} \mathrm{C} ?$ If the $\mathrm{O}_{2}$ gas needed for complete combustion was contained in a $4.75-\mathrm{L}$. flask at $22^{\circ} \mathrm{C},$ what would its pressure be?

The hydrocarbon octane $\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)$ burns to give $\mathrm{CO}_{2}$ and water vapor:
$$
2 \mathrm{C}_{8} \mathrm{H}_{18}(\mathrm{g})+25 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 16 \mathrm{CO}_{2}(\mathrm{g})+18 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})
$$
If a $0.095-\mathrm{g}$ sample of octane burns completely in $\mathrm{O}_{2},$ what will be the pressure of water vapor in a $4.75-\mathrm{L}$. Thask at $30.0^{\circ} \mathrm{C} ?$ If the $\mathrm{O}_{2}$ gas needed for complete combustion was contained in a $4.75-\mathrm{L}$. flask at $22^{\circ} \mathrm{C},$ what would its pressure be?

Key Concepts

Stoichiometry

Stoichiometry involves using the balanced chemical equation to relate the moles of reactants to the moles of products in a chemical reaction. It is key to determining how much of each substance participates in or is produced by a reaction, ensuring that the conservation of atoms is maintained.

Molar Mass and the Mole Concept

The mole is a fundamental unit in chemistry that represents a specific number of particles, and molar mass is used to convert between mass and moles. This conversion is essential for quantifying reactants and products based on sample mass.

Ideal Gas Law

The ideal gas law, expressed as PV = nRT, relates the pressure, volume, number of moles, and temperature of a gas. It is critical for calculating any one of these properties when the others are known, under the assumption that the gas behaves ideally.

Temperature Conversion

Converting temperature from degrees Celsius to Kelvin is necessary for gas law calculations because the ideal gas law requires absolute temperature. This conversion ensures accurate and consistent results in thermodynamic computations.

Reaction Yield and Gas Production

Determining reaction yield involves calculating the amount of product generated from a given amount of reactant using the stoichiometric ratios from the balanced equation. When the products are gases, this concept is combined with the ideal gas law to relate the moles of gas produced to the pressure it exerts in a container.

Transcript

00:01 So in this reaction, we're given a certain amount of octane, and that's going to give us a amount of moles of octane, and then from there, that's where everything, then the whole problem goes from there.

00:10 Since a lot of these species are in the gas form, in fact, all of them are in the gaseous form, we use the ideal gas law pv equals nrt many times, and so we need to have that to work with.

00:22 And so let's start with the amount of octane we're given.

00:25 That's 0 .095 grams.

00:28 Now, to figure out the moles, we're going to use the, molecular weight of octane which is 114 .23 grams per mole and so the grams cancel out the moles will really be in the numerator and that gives us the number of moles that we have and that's equal to 8 .31 times 10th the negative 4th moles of octane.

00:50 Now the first thing we want to deal with is the water vapor.

00:54 So we have for every two moles of octane we have 18 moles of h2o.

00:59 In other words, we have nine times more moles of h2o.

01:02 And so we're going to have to take this number right here, which the moles of octane, go times nine.

01:08 And that is equal to 7 .48 times 10 to negative third moles of h2o as a gas.

01:18 And we're told that the temperature we would have it at is going to be equal to 30 degrees celsius, which is also add 273 to that.

01:26 That's equal to 303 kelvin.

01:30 And this is the form we want to work in.

01:33 Our gas constant are we going to use is 0 .0821 liters, atmospheres per mole kelvin.

01:39 I do this because this is the one i'm most comfortable with...

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