Worked example: Relating reaction stoichiometry and the ideal gas law | AP Chemistry | Khan Academy

So we're told that silver oxide decomposes according to the following equation. 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. I'll give you a little bit of a hint. You're used to saying, "Oh, 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. A little bit of a hint is the ideal gas law. It tells us that ( PV = nRT ). If we solve for ( n ), remember ( n ) is just the number of moles. So, we divide both sides by ( RT ); we get ( n = \frac{PV}{RT} ).

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 to Kelvin. To help you there, I will give you some of the constants and the conversions, so see if you can have a go at this now.

All right, now let's work through this together. The number of moles of oxygen is going to be equal to the pressure of our oxygen, 1.22 atmospheres, times the volume of oxygen, 1.50 liters, divided by the ideal gas constant. We have to use the right one that is going to deal with atmospheres, liters, and Kelvin.

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. This is our ideal gas constant.

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 to the ones place when we convert to Kelvin. 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. 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 \div 0.08206 ) and then we're going to also divide that whole thing by 303. Let's see how many significant figures we have. 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 to produce this volume at this pressure at this temperature.

We're going to need two times this number of moles of silver oxide because notice the two. 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, once again, three significant figures: 0.147.

So, this is approximately 0.14 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 the molar mass of silver times two plus the molar mass of oxygen. Let me get the periodic table of elements out.

The molar mass of silver is 107.87 and oxygen 16.00. So that gives us ( 107.87 ) for each of the silvers, then plus ( 16.00 ) for the oxygen, which gives us ( 107 \times 2 ) equal to that.

Let's see, it goes to the hundredths 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.

If we take the moles of silver oxide and we multiply that times 231.74 grams per mole, 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 because we have three here and five here.

So, we take that and multiply it by ( 0.147 ) to get this, and three significant figures would be 34.1. Approximately 34.1 grams of silver oxide is required to produce this much oxygen.