Consider the reaction:
2 SO₂(g) + O₂(g) → 2 SO₃(g)
b. If 187.2 mL of SO₃ is collected (measured at 315 K and 50.0 mmHg), what is the percent yield for the reaction?

Question

Consider the reaction:

2 SO₂(g) + O₂(g) → 2 SO₃(g)

b. If 187.2 mL of SO₃ is collected (measured at 315 K and 50.0 mmHg), what is the percent yield for the reaction?

Accepted Answer

Hey everyone welcome back. So let's get started with this video. So here they told us to consider the following reaction. And we're told that at 345 Kelvin and 75 mm of Mercury to 65.3 mL of S. 03 were collected. 1 305.2 mL of S. 02 And 176.4 ml of 02 reacted. And they want us to find the percent yield the reaction. Okay, so here we're going to have to use the ideal gas law which is PV is equal to N. R. T. Okay, and since we're looking for the percent yield, we're going to have to solve for moles which is N. So it's going to be N is equal to P B over R. T. So P is pressure. And we're going to have to use atmospheres but were given millimeters of mercury. So we're going to have to convert it volume. Has to be in leaders were given middle leaders. So we're going to have to convert it by dividing by 1000 R. Is a constant T. Is temperature. That has to be in kelvin. They're given to us in kelvin. So we're fine. Okay, so let's start by converting our pressure two atmospheres. So then we have 75 millimeters of mercury and if you recall millimeters mercury in one atmosphere. And we do the math. And we get zero 0.0 Atmospheres. Okay. So with that being said, we can go ahead and figure out the most created when we start with S. 02. So we're going to start with S. 02. So then this is the volume and obviously the pressure is going to be this one temperature is given an R. Is a constant. So then we have P which is zero 0.0 987 atmospheres volume 305.2 million L. We're going to convert it to leaders so we move the decimal back three times and we get zero 3052 leaders Over R which is 0.08206. Leader atmosphere over more kelvin. And the temperature kelvin 3 45 Kelvin which is given to us. So atmospheres cancel out leaders, cancel out kelvin, cancel out and we get moles, We do the math and we get 0. 106 Rose Of S. 02. Okay, so now we're going to use a mole ratio to see how much S. 0. 3 that it creates. So then two moles of S. 02 for every two moles of S. 03. So then two more and so two for every two more. That's so three. So most of us are to cancel out And we get 0.00 small. That's still three Made, starting from s. 0. 2. So let's do the same for the other one which is 02 to see which one of these two is a limiting reactant. So then we start with the same pressure. 0.0987 atmosphere. Okay, the volume is different. It is 176.4 ml. We move the decimal back three times and we get 0. for leaders R is constant 0.08206 Leaders atmospheres over mole kelvin Temperature is the same. 3 45 Kelvin. So then atmospheres cancel out, leaders, cancel out Kelvin cancel. And we get most we do the math and we get 0. 615 Role of 02. Once again we're going to do the mole ratio and we have one mole of 02 for every two moles of S. 03. So one more oh two for two more S 03. More info to cancel out. And here we get 0. 123 Rolls of s. 0. 3. So if you see these the one that creates the least Is the Top one. So that are limiting reactant is going to be S. 02. Okay, so with this limiting reactant now we can find the theoretical yield. So let's take the value of the limiting reactant which is zero point 106 walls S. 0. 3 created from the limiting reactant. And we're going to get the g. And this is a theoretical yield. So we're going to use a molar mass of S. 03. So then one mole of S. 03 is equal 2 80 g. So then moles cancel out And we get zero points 04, sorry, eight four eight g of s. And this is our theoretical. Okay, so now what we need is the actual yield from the experiment. And if we go up we are told that and you get a different color. 265.3 mm of S 03 were collected. So this is an actual yield from the experiment. And the question. So then we're going to take those middle leaders and get grams And how are we going to do that? We're going to get the molds using the PV equals NRT equation. And then use the molar mass to get the grounds. So then let's go back down. So here we're going to use the same one and is equal to PV over R. T. Once again, Pressure is going to be the same one. Zero point 0987 Atmospheres. The volume. This is going to be the one that they told us they collected which was 265. millimeters. We move the decimal back three times or divide by 1000 And we get zero 265 three leaders Are the constant 0.08-06 L per atmosphere. Sorry leader times atmosphere over more kelvin. The temperature is the same. 3 45 Children. So the temperature for kelvin cancel out atmosphere, cancel out leaders. And we get more And we get 0. zero 9-5 malls Of S. 0. 3. Okay now we're going to get the grounds by using the molar mass. one more of s. 0. 3 Is equal to 80 g. We do the math and we get zero point 0740g S. 03. And this is from the experiment from the question. So this is going to be the actual. Okay. And if you recall percent yield is equal to actual over theoretical times 100. We have actual we have theoretical and then we just plug it into this. So the actual is 0.07 Over theoretical 0.0848 Times 100. So here grabs would cancel out and we would get 87 point 26%. Okay. Thank you for watching. I hope this helps. And I'll see you in the next video

Consider the reaction:2 SO2(g) + O2(g) → 2 SO3(g)b. If 187.2 mL of SO3 is collected (measured at 315 K and 50.0 mmHg), what is the percent yield for the reaction?

Verified Solution

Consider the reaction:
2 SO₂(g) + O₂(g) → 2 SO₃(g)
b. If 187.2 mL of SO₃ is collected (measured at 315 K and 50.0 mmHg), what is the percent yield for the reaction?