Carbon monoxide gas reacts with hydrogen gas to form methanol. CO(g) + 2 H₂(g) → CH₃OH(g) A 1.50-L reaction vessel, initially at 305 K, contains carbon monoxide gas at a partial pressure of 232 mmHg and hydrogen gas at a partial pressure of 397 mmHg. Identify the limiting reactant. Determine the theoretical yield of methanol in grams.

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Welcome back everyone to another video, carbon monoxide gas reacts with hydrogen gas to form methanol. We're given the balanced equation. A 1.50 L reaction vessel initially at 305. Kelvin contains carbon monoxide gas at a partial pressure of 232 millimeters. Mercury and a high gas at a partial pressure of 397 millimeters. Mercury identify the limiting reactant and determine the theoretical yield of methanol in grams. And we're given four answer choices. We are going to solve this problem. First of all, let's recall the ideal gas law which states that the product of pressure and volume should be equal to the number of moles and multiplied by the universal gas constant R and multiplied by the absolute temperature T. So if we want to identify the limiting reactant for the first, for the first part of the problem we essentially want to solve for moles, which is the ratio between PV and RT. Now let's apply this formula. First of all, we'll start with the number of moles of carbon monoxide. We're going to use the partial pressure. In this case, if we carefully look at the given data, we have 232 millimeters mercury. So we're going to include that, we're going to convert that into atmospheres. We can use our conversion factor. We will include millimeters mercury on the bottom and atmosphere suns. So now what we know is that one atmosphere is equivalent to 760 millimeters mercury. And we also need to use our volume. Let's include volume. In this case, that's a 1.5 L. And now we need to divide by RT. So what is our universal gas constant? Well, essentially that would be 0.08206 L multiplied by atmospheres divided by mole multiplied by Kelvin. And finally, we are going to multiply this number by the absolute temperature. In this case, that would be 305 Kelvin. And now what we can do, we can essentially apply the same formula and evaluate the number of moles of hydrogen gas, right? So let's do that. We will need to replace the pressure, right? And now what else every other parameter remains the same? So essentially what we're going to do here, we're going to and the pressure. So that'll be 397 right? And everything else stays the same. So from here, we can essentially state that the number of moles of carbon monoxide would be equal to zero point. Let's go ahead and write that number 0.018 295 moles. Now, if we look at the results for the number of modes of hygen, we find out that this number is equal to 0.0313 07 miles. Now, the question is which one of them is our limiting reactant? Well, we noticed that the ratio of carbon monoxide to hygen is 1 to 2 according to the balanced equation. And what we want to do, we want to essentially divide these number of modes by sty metrical efficients and find out which one is the lower number. So first of all, we need to divide the number of moles of co by one because that's the story geometric efficient in the balanced equation. And we want to divide the number of moles of hydrogen by two. And clearly if we do that, we get our equivalences, let's see what we get for the number of moles of carbon monoxide. Nothing really changes. We get an CEO divided by one that will be equal to the same number of 0.018295. And if we take the number of moles of hydrogen and divide that by two, because that's our coefficient. So we're going to essentially include our subscript properly. And now what do we get if we divide 0.031307 by two? Well, essentially let's calculate that the result that we get should be 0.0156 54 moles. So now the question is which one of them is lower. We noticed that the second number is lower. So hydrogen gas is the limiting reactant. We are going to use the shorthand notation. We're going to say that this is our LR and now we want to find the theoretically yield of methanol, right. So let's go ahead and do that. If we carefully look at the balanced equation, we notice that two moles of hydrogen produce one mole of methanol. And what we're going to do here, we're going to use that fact and identify the number of moles of methanol C 30 according to ST geometry, that will be half of the number of moles of hydrogen, right? We already have that number from our previous calculation, but that would be 0.015654 moles. And now if we want to convert that into mass, we simply want to use the molar mass of methanol. And we're going to use 0.015 654 moles. We need the molar mass of Masol. So we can essentially use the periodic table for that purpose. The molar mass of methanol would be 32.04. And eventually we get the theoretical yield of methanol multiplying these numbers together. We can say that we end up with 0.502 g. So essentially, we can now conclude our answer if we go back to the answer choices, we have four of them. And we essential want to see which one is the correct one. First of all, let's remember that our limiting reactant is hygen and the theoretical yield of methanol is 0.502 g. Therefore, we can conclude that option B is the correct option and the correct answer to this problem. Thank you for watching and I hope to see you in the next one.

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Key Concepts

Limiting Reactant

Partial Pressure

Theoretical Yield