### Video Transcript

Propane undergoes a combustion reaction according to the following equation. C₃H₈ plus 5O₂ react to form 3CO₂ plus 4H₂O. If 33.6 liters of carbon dioxide gas is produced, how many grams of propane gas are burned, assuming STP? A) Seven grams, B) 11 grams, C) 20 grams, D) 22 grams, or E) 66 grams.

STP stands for standard temperature and pressure. Standard temperature and pressure conditions are zero degrees Celsius and one bar. One bar is very close to one atmosphere, with one atmosphere equal to 1.01325 bar. What we’re dealing with here is the combustion reaction, the reaction with oxygen of propane. In the chemical equation, we only have carbon dioxide and water, meaning that we’re dealing with a complete combustion reaction. At standard temperature and pressure, propane is a gas. Propane’s boiling point is well below zero degrees C, at negative 42 degrees Celsius.

The question tells us that this reaction is producing 33.6 liters of carbon dioxide gas. And our job is to figure out the mass of propane gas that was burned to begin with. In the scenario, we’re burning a mixture of propane and oxygen. And we’re measuring the volume of pure carbon dioxide gas we get out. Now, we can use the ideal gas law to work out the amount in moles of carbon dioxide gas present in 33.6 liters. But we’ve already been told we’re dealing with standard temperature and pressure. So there’s a shortcut.

At standard temperature and pressure, there’s a fixed relationship between the volume of the gas and the amount of gas if it’s ideal. From this relationship, we get the molar volume, 𝑉 m. This value tells us the number of liters for each mole of gas we can expect. And the molar volume for an ideal gas at zero degrees Celsius and one bar is 22.4 liters per mole. So at standard temperature and pressure, if we measure the volume of a pure gas to be 22.4 litres, we know we have one mole of that gas. If we have 44.8 litres of it, we have two moles, regardless of the identity of the gas. So we can work out the number of moles of carbon dioxide gas in our 33.6 liters by multiplying 33.6 liters by one mole per 22.4 liters. This gives us 1.5 moles of carbon dioxide. But the question asked us to find the mass of propane. So we need to convert moles of carbon dioxide to moles of propane. The ratio of carbon dioxide molecules produced in the reaction to propane molecules consumed in the reaction is three to one.

To work out the number of moles of propane, we can take the number of moles of carbon dioxide produced and multiply it by one mole of propane per three moles of carbon dioxide. This means that, in the reaction that produces 33.6 liters of carbon dioxide, we must have used up 0.5 moles of propane. Now, all that remains is to work out the mass of 0.5 moles of propane. In order to do that, we’re going to need to work out the molar mass of propane. To do that, we can look at the atomic masses of the elements that make up propane.

There are three carbons in propane. So they each contribute 12 grams per mole to the molar mass of propane. We also have eight hydrogens per propane molecule. So each hydrogen contributes one gram per mole to the molar mass of propane. If we sum the contributions together, we end up with a molar mass of propane of 44 grams per mole. We can use the rounded values for this calculation because we’re only looking for a value accurate to one or two significant figures.

We can use the molar mass and the amount in moles of propane to work out the mass of our sample. 0.5 moles of propane multiplied by 44 grams per mole is equal to 22 grams of propane. What this means is that if we were to burn completely 22 grams of propane, separate out the carbon dioxide, and have it stored under standard temperature and pressure. The volume of carbon dioxide we would get will be 33.6 liters.

So if we got 33.6 liters of carbon dioxide gas from burning propane, the mass of propane we must have burned, assuming STP, would’ve been 22 grams.