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## AP®︎/College Chemistry

### Course: AP®︎/College Chemistry>Unit 4

Lesson 4: Stoichiometry

# Worked example: Relating reaction stoichiometry and the ideal gas law

AP.Chem:
SPQ‑4 (EU)
,
SPQ‑4.A (LO)
,
SPQ‑4.A.1 (EK)
,
SPQ‑4.A.2 (EK)
,
SPQ‑4.A.3 (EK)
By combining stoichiometric calculations with the ideal gas law, we can calculate amounts of reactants and products for chemical reactions involving gases. Created by Sal Khan.

## Video transcript

- [Instructor] So we're told that silver oxide decomposes according to the following equation. So for every two moles of silver oxide it decomposes into four moles of silver and one mole of molecular oxygen. How many grams of silver oxide are required to produce 1.50 liters of oxygen gas at 1.22 atmospheres and 30 degrees Celsius? So I want you to pause this video and see if you can figure this out. And I'll give you a little bit of a hint. So you're used to saying, well, if I know how many moles of oxygen I need to produce, then I need twice as many moles of silver oxide because the ratio of moles of silver oxide to molecular oxygen is two to one. But they don't tell us how many moles of molecular oxygen we're producing. They give us volume and pressure and temperature. But a little bit of a hint is the ideal gas law. It tells us that PV is equal to NRT. And if we solve for N, remember N is just the number of moles. So we divide both sides by RT. We get PV over RT is equal to N. And it looks like they gave us all of this stuff. The pressure is right over here. The volume is right over here. The ideal gas constant, we can look that up, and then temperature, we just have to convert this right over here into Kelvin. And to help you there, I will give you some of the constants and the conversions. And so see if you can have a go at this now. All right, now let's work through this together. So the number of moles of oxygen is going to be equal to the pressure of our oxygen. So 1.22 atmospheres times the volume of oxygen, times 1.50 liters divided by the ideal gas constant. And we have to use the right one that is going to deal with atmospheres, liters and Kelvin. So if we're dealing with atmospheres, liters and Kelvin, then we'll use this ideal gas constant right over here. So divided by 0.08206. I'll write the units here. Liters, atmospheres, divided by moles, and also divided by Kelvin. So this is our ideal gas constant, and then we're going to multiply that times the temperature in Kelvin. Now they only gave us two significant figures here. They're only going to the ones place. So let's only go only go to the ones place when we convert to Kelvin. So let's just add 273 to this. So this is going to be times 303 Kelvin. And let's make sure the units work out. That cancels with that, that cancels with that, that cancels with that. And if we have a dividing by moles in the denominator, then that's just going to become a moles eventually in the numerator. So this is going to be approximately equal to 1.22 times 1.5 divided by 0.08206. And then we're going to also divide that whole thing by 303. Is equal to this. And let's see how many significant figures. We have three, we have three, we have a lot more than three right over here. And then we have three right over here. So I'm going to round to three significant figures. So 0.0736. So 0.0736 moles of molecular oxygen is how many we need, how much we need to produce to get this volume at this pressure at this temperature. And so we're going to need two times this number of moles of silver oxide, because notice for every one mole of molecular oxygen we produce we need twice as many moles of silver oxide. So let's multiply this times two. So, times two gets us, and once again, three significant figures, 0.147. So this is approximately 0.147 moles of silver oxide that we need to produce. But they're not asking us how many moles of silver oxide are needed. They're asking us how many grams. So we just have to multiply this times the molar mass. So let's first figure out the molar mass of silver oxide. I'll write it right over here. So silver oxide's molar mass is going to be whatever the molar mass of silver is times two, plus the molar mass of oxygen. And so let me get the periodic table of elements out. Molar mass of silver is 107.87, oxygen, 16.00. So that gives us 107.87 for each of the silvers. And then plus 16.00 for the oxygen, which gives us 107.87 times two is equal to that. Let's see it goes to the hundredth place. And then plus 16.00 also goes to the hundreds place. So we'll go to 231.74. Also going to the hundreds place. I'm just keeping track of the significant figures when we're adding. So this is 231.74 grams per mole of silver oxide. And so if we take the moles of silver oxide and we multiply that times 231.74 grams per mole, and notice the moles cancel out, and we're just left with the grams, which is what we want. This is going to be approximately equal to, and we're going to end up with three significant figures 'cause we have three here and five here. So we're going to take that and multiply it by 0.147. Gives us this. And three significant figures would be 34.1. Approximately 34.1 grams of silver oxide is required to produce this much oxygen.