The flame produced by the burner of a gas (propane) grill is...

The flame produced by the burner of a gas (propane) grill is a pale blue color when enough air mixes with the propane $\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)$ to burn it completely. For every gram of propane that flows through the burner, what volume of air is needed to burn it completely? Assume that the temperature of the burner is $200^{\circ} \mathrm{C},$ the pressure is 1.00 atm, and the mole fraction of $\mathrm{O}_{2}$ in air is 0.21.

Key Concepts

Ideal Gas Law and Molar Volume

The Ideal Gas Law (PV = nRT) relates the pressure, volume, number of moles, and temperature of an ideal gas. It is used to calculate the molar volume—the volume occupied by one mole of gas under specific conditions—which is essential when converting between moles and volume in gas-phase reactions.

Air Composition and Mole Fraction

Air is a mixture primarily composed of nitrogen and oxygen, with oxygen typically constituting a smaller fraction by volume. The mole fraction of oxygen in air is used to determine how much total air is needed to provide a specific number of moles of oxygen for a reaction.

Reaction Stoichiometry

Reaction stoichiometry involves using balanced chemical equations to determine the relationships between the amounts of reactants and products. It allows the conversion from a given mass of a substance to the number of moles and determines how many moles of another reactant are required, based on their coefficients in the balanced equation.

Combustion Reactions

Combustion reactions are exothermic chemical processes in which a fuel reacts with an oxidizer (often oxygen) to produce oxidized products such as carbon dioxide and water. Understanding these reactions is crucial for calculating the necessary amount of oxidizer for complete fuel combustion.

Transcript

00:01 We want to know what volume of air is required to completely combust one gram of propane.

00:07 We know that for combustion reactions, whichever hydrocarbon we start with, and it would be propane in this case, has to react with oxygen, and the combustion products are always carbon dioxide, co2, and water, h2.

00:23 So the first step that we have to take in this problem is to balance this combustion reaction.

00:29 We start by looking at the number of carbon atoms on each side of the reaction, since carbon only appears once on the left and right side of this equation.

00:41 We can see that if we place a one here, then there are three carbon atoms on the left side, and so this stoichiometric coefficient would have to be a three in order to balance out the carbons.

00:55 Next, we see that hydrogen only appears once on the left and the right side of the equation.

01:02 And since this stoichiometric coefficient is a one, there must be eight hydrogens on the left side of the reaction.

01:12 And so in order to balance that out, there have to be four moles of water.

01:19 So that one multiplied with the two hydrogen atoms from water, there are a total of eight hydrogen atoms on either side of the reaction.

01:30 So now all that leaves is the number of oxygen atoms.

01:33 When we look, we see that this stoichiometric coefficient of three, when multiplied with the two oxygen atoms from co2, results in six oxygen atoms.

01:44 And when added to four moles of hydrogen times one oxygen atom for each mole of water, we get a total of 10 oxygen atoms on the the product side.

01:58 And so this stoichiometric coefficient would have to be a five so that when it's multiplied with two oxygen atoms from o2, there are also 10 oxygen atoms on the left side.

02:10 So that is a balanced stoichiometric combustion reaction for propane.

02:15 We are given a temperature and a pressure when we want to determine the volume of air required to combust one gram of propane.

02:22 We need to use the ideal gas law in order to solve for the volume of air.

02:28 So that v equals n rt over p.

02:35 We are given values for pressure and temperature, and we know the ideal gas constant...

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