- Worked example: Calculating amounts of reactants and products
- Limiting reactant and reaction yields
- Worked example: Calculating the amount of product formed from a limiting reactant
- Worked example: Relating reaction stoichiometry and the ideal gas law
Worked example: Relating reaction stoichiometry and the ideal gas law
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.
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- Sal, the number of times you used that calculator 😭 wish I could do so in an exam(3 votes)
- It is possible to do math without a calculator, people have been doing so for thousands of years. Calculators just make arithmetic faster.(3 votes)
- In Germany, we use the metric system.
Therefore, I believe my equation is different.
Would it be easier in my case just to use the US system(Metric) rather than Germany(Imperial)??
- THE WATCHER(1 vote)
- All the units used here are SI units, of which the metric system is a subset of. So essentially if they're using SI units, they're using the metric system as well. So if you're accustomed to using the metric system, all the equations and units used in the video should be useable.
As a side note, here in US we use US customary units while the British use imperial units. They are similar in unit names and values, but are technically different. In any case, I wouldn't use either US customary or imperial units for scientific calculations. They're more difficult to convert and it's harder to communicate data with the international scientific community who use SI units.
Hope that helps.(5 votes)
- What is PV=nRT? This wasn't mentioned in any of the previous lessons. I wish this course wasn't so messy.(3 votes)
- That's the ideal gas law. Sal mentions this at0:50. It relates an ideal gas's pressure, volume, number of moles, and temperature to each other.
In the AP chemistry section the ideal gas law comes before this lesson. https://www.khanacademy.org/science/ap-chemistry-beta/x2eef969c74e0d802:intermolecular-forces-and-properties/x2eef969c74e0d802:ideal-gas-law/v/ideal-gas-equation-pv-nrt
Hope that helps.(1 vote)
- At3:40how is it molecular oxygen? The question does not specify its diatomic oxygen so shouldn't we assume its just 1 atom of oxygen?(1 vote)
- We can assume that they mean molecular oxygen first because that is one of the products in the chemical equation that they give in problem.
Secondly we can assume molecular oxygen because they ask for how much oxygen gas is produced. Oxygen gas's most stable allotrope (or form) at room temperature and pressure is diatomic oxygen.
In general if the problem involves oxygen or oxygen gas, we assume that it is diatomic oxygen unless otherwise stated.
Hope that helps.(2 votes)
- Does temp and preassure really matter that much in stoich?(1 vote)
- When dealing with gases definitely. Different atmospheric or gas pressures and surrounding temperatures can affect the amount of gas that can be present in a specific volume. The partial pressure of a gas can also be used to represent the concentration of a gas relative to other gases in a system and is commonly used in Kp expressions and even in rate laws instead of cocentrations. Also, when not enough info is given to solve for moles, you can sub in the pressure of the gas of the volume to do stoich since both of them are directly proportional to moles of gas(2 votes)
- why #of Ag2O mole is two times the #of O2 mole?
Answer= bwcause we know O2 is limiting reactant since it has smallest coefficient
am i right?(1 vote)
- There are twice as many moles of silver oxide as there are oxygen gas because the ratio of silver oxide to oxygen is 2:1, which we know from the coefficients of the chemical reaction. Sal explains this at0:32.
Your answer has several mistakes. Firstly, this isn't a limiting reactant problem. We're trying to know how much reactant is required to make a certain amount of product. Second, oxygen gas isn't a reactant, it's a product. Third, we don't determine limiting reactants based on their coefficients.
Hope that helps.(1 vote)
- at3:11you shouldn't multiply by 303k instead of dividing ?(1 vote)
- That would be like calculating ((1.22)(1.50)(303))/(0.08206), which would be a different answer.(1 vote)
- I wasn't prepared for atm pressure(1 vote)
- where does the 303 K come from when calculating n?(1 vote)
- The formula for finding Kelvin from Celsius is:
T_(K) = T_(°C) + 273.15
In this example it was just rounded to a whole number.(1 vote)
- Why do you have to use the ideal gas law to solve this problem? Can't we take the number of grams in 1.50 L, divide it by 15.99*2 to get the number of moles, and then multiply it by 2 (from the 2:1 ratio of Ag(2)O and O(2)), then multiply by the atomic mass of Ag(2)O to get the number of grams needed?(0 votes)
- Well, how would you get the number of grams of oxygen gas from the volume though? You could use density, but that can change due to pressure and temperature. So you'd have to find the density for oxygen gas at 1.22 atm and 303K.(2 votes)
- [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.