Gasoline, which we will represent as having the properties...

Gasoline, which we will represent as having the properties of isooctane, is consumed by an idling automobile engine at a rate of 1 gal/h. A monitor in the garage where work is being done on the engine detects an accumulation of $\mathrm{CO}$, indicating incomplete combustion of the gasoline. What does this information imply about the gasoline-to-air ratio being fed to the engine? If we assume that gasoline has the properties of isooctane, $\mathrm{C}_{8} \mathrm{H}_{18}$, estimate the feed rate (mol/h) of air for $10 \%$ excess oxygen fed to the engine?

Key Concepts

Stoichiometric Combustion

This is the ideal combustion process in which a fuel reacts completely with oxygen in exactly the right proportions, so that all the fuel is burned without any excess air and no carbon monoxide or unburnt hydrocarbons are formed. It is a fundamental concept for determining the precise amount of oxygen required for the complete combustion of a given fuel, and it provides a baseline against which mixtures with excess or insufficient air are evaluated.

Incomplete Combustion

Incomplete combustion occurs when there is not enough oxygen to allow the fuel to react completely to form carbon dioxide and water. As a result, products such as carbon monoxide, soot, or other partially oxidized compounds are produced. This concept is important in diagnosing engine performance issues and in controlling environmental pollution due to toxic byproducts.

Air-Fuel Ratio

The air-fuel ratio is the mass or molar ratio of air to fuel that is introduced into an engine for combustion. It is a critical parameter that affects efficiency, emissions, and performance. A precise air-fuel ratio based on stoichiometric combustion ensures complete fuel burning, while deviations can lead to incomplete combustion or suboptimal engine operation, often seen in cases with excessive or insufficient oxygen.

Combustion Reaction Stoichiometry

This concept involves balancing chemical equations for the combustion of hydrocarbons, allowing for the calculation of the amounts of reactants (fuel and oxygen) and products (carbon dioxide and water). By applying stoichiometry, one can determine the theoretical air requirements based on the chemical composition of the fuel, which is essential for designing combustion systems and for troubleshooting issues like excess emissions.

Excess Oxygen in Combustion

Excess oxygen refers to the deliberate addition of more oxygen than what is theoretically required for stoichiometric combustion. This is commonly done to ensure complete combustion, reducing the formation of harmful byproducts such as carbon monoxide. Understanding the impact of excess oxygen is vital when adjusting engine settings to optimize performance and minimize emissions.

Transcript

00:04 So the first question is asking us to write a balanced equation for the combustion of octane.

00:13 So combustion reactions are always going to have oxygen gas, and they're going to combust into co2 and h2o.

00:25 And so if we go ahead and try to balance this out, well, we can see we've got eight carbons on the left, which means that we need eight carbons on the right as well.

00:34 We've got 18 hydrogens on the left.

00:37 So on the right, we need 18 hydrogens as well, which means that we need nine waters.

00:43 And therefore, the number of oxygens we need, so the oxygen's on the right.

00:47 We've got 8 times 2 is 16, plus 9 more is 25.

00:53 So we're going to need 25 over 2 oxygens in order to balance this out.

01:02 So that's a balanced equation.

01:03 Ideally, i would multiply all of those through by two as well, because we don't like to have non -intigures.

01:11 So we'd end up with a 2, 25, 16, and 18.

01:24 So part 2, part b, is saying that if we assume that gasoline is 100 % isoctane, and the density of isoctane is 0 .72 grams per milliliter.

01:47 Then the question is, what mass of co2 in kilograms is produced by burning 4 .6 times 10 to the 10th liters.

02:03 So that's 4 .6 times 10 to the 10th liters.

02:13 And the question is, how many grams of co2 do we create? so what we're going to need to do here is let's convert our .792.

02:31 Actually, let's convert the liters here into milliliters.

02:35 So we're going to do 10 to the 13th milliliters.

02:42 So 0 .792 grams per milliliter times 4 .6 times 10 to the 13th milliliters.

02:53 And this is to figure out how many grams of c8h18 we have.

03:01 So let's multiply this out.

03:04 792 times 4 .6 times 10 to the 13th power gives us 3 .64 times 10 to the 13th grams of c8.

03:35 H18.

03:40 So that's not what the question is asking though.

03:42 The question is asking how many grams of co2 do we get out of this? so we're going to need to do some stoic here.

03:51 3 .64 times 10 to the 13th grams of c8h18.

04:04 And the molar mass of that 12 times 8 is 96 plus 18...

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