A steel container with a volume of 500.0 mL is evacuated,
and 25...

Question

A steel container with a volume of 500.0 mL is evacuated, 
and 25.0 g of CaCO3 is added. The container and contents 
are then heated to 1500 K, causing the CaCO3 to decompose 
completely, according to the equation 
CaCO31s2¡CaO1s2 + CO21g2. 
(b) Now make a more accurate calculation of the pressure 
inside the container. Take into account the volume of 
solid CaO 1density = 3.34 g/mL2 in the container, and 
use the van der Waals equation to calculate the pressure. 
The van der Waals constants for CO21g2 are 
a = 3.59 1L2 # atm2>mol2 and b = 0.0427 L>mol.

Accepted Answer

Hi everyone. This problem reads a 15 g sample of magnesium carbonate is placed in a one liter reaction vessel along with excess nitric acid. The following reaction then occurs at 350 Kelvin. Use the Van der Waals equation to calculate the pressure of carbon dioxide gas in the vessel, taking into account the volume of the non gaseous products. The constants for carbon dioxide gas are A and B equal the following. Assume the non gaseous products occupy a total volume of 111.6 mL. So, our goal for this problem is to calculate the pressure of carbon dioxide gas in the vessel and were told to do it using the Van der Waals equation. So let's go ahead and write down that equation. It's P is equal to N R T over V minus N b minus a N squared over V squared. So, this is our Van der Waals equation to calculate pressure. Alright, So Let's go ahead and take a look at what we're given and what we're missing. So let's start off with our variable end and represents moles. Okay, and so we're not directly given moles in this problem. So, we need to calculate it. We're told that we have 15 Let's do this in a different colour. We're told that we have 15 g of magnesium carbonate. Okay, So what we want to do is we want to go from grams of magnesium carbonate, two moles of carbon dioxide, because the question is asking us to calculate the pressure of carbon dioxide. So, this is our goal here and we're going to go ahead and start off with what we're given. So we want to go from grams of magnesium carbonate, two moles of magnesium carbonate. And we'll do that using the molar mass for magnesium carbonate. So in one mole of magnesium carbonate, the molar mass is using the periodic table 84.3139 g of magnesium carbonate. Alright, so making sure our units cancel. Here, we see that grams of magnesium carbonate cancel. And we're left with moles of magnesium carbonate. Now we want to go from moles of magnesium carbonate, two moles of carbon dioxide. So let's look at our reactions. So we can see what the multi mole ratio is. Okay, so for every one mole of magnesium carbonate consumed, we see that one mole of carbon dioxide is produced. So we have a 1-1 mole ratio. So we'll go ahead and use that here. So one mole of magnesium carbonate for every one mole of carbon dioxide. So our moles of magnesium carbonate cancel. And we're left with moles of carbon dioxide, which is what we're looking for. So now we can go ahead and do our calculation to solve four moles of carbon dioxide, once we do it, we get 0. 99 moles of carbon dioxide. Okay, so now we know what n is our represents our gas constant. So let's go ahead and write them down as we figure it out. So we know N is equal to 0. 99 moles of carbon dioxide are represents a gas constant that we should have memorized. That is 0.08206 leaders. Atmosphere over Mo Calvin T represents temperature and the temperature is given in the problem and it is 350 Kelvin, which is the exact unit that we need. Okay, V is volume. So let's move this down a little bit. So v represents volume. So the last statement in this problem tells us to assume the non gaseous products occupy a total volume of 111. ml. So we are starting volume is one leader. So we're going to subtract the total volume. So we have 111.6 mL. We're going to convert that to Leaders. So we're going to move the decimal place three units. Okay, so we're going to have one leader minus 0.116 leaders. Gives us a volume of 0. leaders. Okay, A.R. Wal's constant is given in the problem. It is 3. atmospheres leader squared over moles squared. And r Vander wal's constant B is also given in the problem. It is 0. liters over. All Right. So now we have everything that we need to plug into? Our Vander wal's equation to solve for pressure. So let's go ahead and do that referencing all of our givens that we just wrote down. All right, So let's go ahead and plug in. So pressure is going to equal N. So let's just make this a little bit smaller. So we have some space. So N is our moles. So we have 0. moles of carbon dioxide times are 0.08206 leaders atmosphere over Mole Kelvin times, temperature 350 Calvin. And this is all over Be our volume is 0. leaders -N. So this is N is our number of moles. So 0. moles N times B R. Vander wal's constant B is 0. 286 leaders over mall. So that's volume minus N. B. And then we're going to minus Vander wal's constant A Is 3.6. Let's move this down. So we have some space. Sir. Vander was constant A is 3. atmospheres leaders squared over moles squared. So eight times N squared or number of moles squared. So 0.1799 moles squared and this is over volume squared. So the volume is 0.8884 liters squared. So we've plugged everything in for pressure. So now we just need to do the calculation once we do this calculation we get pressure Is equal to 5.66 atmospheres, and this is going to be our final answer. So the using the guy vander Waals equation the pressure of carbon dioxide gas in the vessel, taking into account the volume of the non gasses products is 5. atmospheres. That's it for this problem, I hope this was helpful.

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