When 10.0 g of a mixture of Ca1ClO322 and
Ca1ClO22 is heated to ...

Question

When 10.0 g of a mixture of Ca1ClO322 and 
Ca1ClO22 is heated to 700 °C in a 10.0-L vessel, both compounds 
decompose, forming O21g2 and CaCl21s2. The final 
pressure inside the vessel is 1.00 atm. 
(b) What is the mass of each compound in the original 
mixture?

Accepted Answer

Hi everyone. This problem reads a mixture of barium bromate and barium bro might weighing eight g was placed in a five liter reaction vessel. The vessel was then heated to 800 degrees Celsius and both compounds decomposed forming barium bromide and oxygen gas. If the vessel has a final pressure of 8000.8968 atmospheres, calculate the mass of each compound before decomposition. So we want the mass of each compound before decomposition. So we're looking for the mass of barium bromate and the mass of barium bro might. Okay, so let's start off by writing the decomposition reactions balanced. So we have barium bromate is going to decompose into barium bromide and oxygen gas and we have barry umbro might is going to decompose and to barium bromide and oxygen gas. So these are the balanced reactions. Okay, so because we're dealing with gasses, we're going to use the ideal gas law to calculate the total moles. And let's recall that the ideal gas law is PV equals N. R. T. And because we want to calculate the total moles, we want to isolate the variable moles. So we're going to divide both sides by R. T. And we get N. Is equal to P. V. Over R. T. So let's take a look at our problem to see what we have already. So P represents pressure and our pressure, we're told is 0. atmospheres V. Represents volume and we're told that the volume is five liters R. Is a gas constant 0.8 to 06 leaders atmosphere over mole kelvin and t represents temperature and units Calvin. So we're told the temperature is 800 degrees Celsius and we need this temperature in kelvin. So we're going to add 273.15. So we get our temperature is equal to 73.15 Kelvin. Okay, so we have all of the information that we need to calculate the total moles. So let's go ahead and plug in these values. Okay, so we get total moles is going to equal pressure times volume. So we have 0. Atmospheres Times five l over gas constant R 0. to 06 leaders atmosphere over moles, kelvin times temperature 1 1073. kelvin. So this is everything that we need to plug in. So let's go ahead and do the calculation. And when we do that, we get the total moles is equal to zero point 05092 moles of oxygen gas. Okay, Because the oxygen gas is our gas and our decomposition reaction. Okay, so now that we know what our total moles is, let's go ahead and right out the molar mass for both of our, our barium bromate and our barium burrow might. Okay, because we're going to need that. Okay, so up here, let's go ahead and just make a note. So our molar mass for barium bromate is equal to . grams per mole. And our molar mass for barry and bro might. That is equal to 361. g per mole. Alright, so now that we know what the total moles is, because we have the mixture, we're going to set this mixture equal to the total weight of the mixture. So the problem tells us a mixture of barium bromate and barium bro might weighs eight g. Okay so we're going to take eight g which is the total mass of our mixture. And we're gonna set that equal to the mass of barium bromate plus the mass of barium bro might. Okay so what essentially this is going to mean is our eight g is going to equal? We're going to need the moles. Okay so we have massive barry and bro might. So we're going to take the molds of barry and bro might and multiply that by the molar mass of barry and bro, might because that's going to give us the moles and we're going to do the same thing for barium bromate for bro. Might. Excuse me. So we're going to take the moles of barry and bro might and multiply it by the molar mass of barium bromate. Make this smaller. Okay, so now we're going to do some substitution. So we're going to let X equal the moles of barium bromate and we're going to let y equal the moles of barium bro might. Okay, because these are the values that we don't know. So we're gonna let X and Y reflect those numbers. Okay, So if we take a look at our problem and we look at the multiple ratios for the moles of oxygen gas, we're going to look at our balanced equation for a moles of oxygen gas for our first decomposition reaction, we see it's going to be three times the malls of barium bromate. Okay, so what that means is that for our first decomposition reaction are moles of oxygen gas is going to equal three times X. Because we said X is the moles of barium bromate. So for the same thing for a second decomposition reaction, the multiple ratio we see we have for moles of oxygen gas, it's going to be two times the moles of barium grow might. Okay, so when we write that out, we see that it's moles of oxygen gas equals to y. Okay, so what we're going to do is we're going to set three X and two Y three X plus two Y. We're going to set the sum of that equal to our total moles and we know what our total moles is because we calculated that so we're going to take three X plus two Y is equal to our total moles, which is 0. moles. Alright, so we're going to solve for y here. So when we solve for y, we're gonna say two, Y is equal to 0. moles minus three X. And we're solving for Y. So let's go ahead and divide both sides by two. And when we do that we get why is equal to 0. moles -3 x. Over to. Okay, so now this is the equation that we're going to use to substitute all of our values in. Okay. So we don't know what X and Y are but we know what the what we are. We were going to substitute the values that the values for the molar masses. Okay, so what we're gonna do is we're going to take The eight g. So we have eight. That this initial equation here. Okay, this initial equation we're going to go ahead and substitute Excuse me, it's this one the second one. This initial equation here, we're going to go ahead and substitute the molar masses in so that we can solve for X and Y. So that's the equation that I'm going to copy down below. So we have eight is equal to X times the molar mass which is 393.15. And this is going to be plus Y times 361.135. Okay, so we have eight Which is the total amount of g is equal to X Times 393. plus. And we said why remember we we said what y is equal to? We said it is this right here. Alright, So we're going to substitute that value in all right? So it's going to be plus 0. moles -3 x divided by two. Okay. And we're going to multiply that By 361.135. So we're just plugging everything in here. Alright, so let's go ahead and simplify. So eight is equal to .135 x plus 0. moles minus three X Times 180.5675. Okay, so we're going to simplify further eight is equal to 393. x plus 9. minus 541. X. So we're going to bring all the excess to one side. So we have 541.7025 x -393. X is equal to 9. -8. Okay, Simplifying 148.  aN:aN:000NaN 148. X is equal to 1.19458. And we're solving for x here. So let's go ahead and divide both sides by 148.5675. And when we do that, we get X is equal to 8. times 10 to the negative three moles. Okay, So now that we know what X is, we're going to solve for Y because we know up above that Y is equal to this. Alright, So let's go ahead and solve for Y by plugging in X. So why is equal to let's do it in red? Since it's in red up above Y is equal to 0. moles minus three times X. Which we just saw for X. X is eight. Yes, 8.401 times 10 to the negative three moles. And this is divided by two. Okay, so once we do that, we get y is equal to 0.0134 moles. So now we know what the molds of X is and we know what the moles of. Why is the question asked us to calculate the mass of each compound. So we need this amount of moles and grams. So we're going to go from mass or we're going to go from moles to grams. So let's go ahead and start off with our mass of barium bromate. So we're gonna go from moles. two g. All right, so our mass of barium bromate. We're going to take the moles. So our moles is 8.401 times 10 to the negative three moles. So we need the molar mass to go from moles to grams. And we wrote it up at the top. We said that in one mole of barium bromate, the mass is 393. g. Making sure units cancel most cancels and we have the mass is equal to 3. g. Okay, so we'll go ahead and write barium bromate Equals 3.16 g. So we'll highlight this because this is one of our final answers. So now let's go ahead and calculate the massive barium bro. Might. Okay, so our mass of barium bro, might We have the moles is equal to 0.0134 moles. Okay, remember that X and Y represented our moles X was our mass for barium bromate and y is our mass for barium bro. Might. So the 0.134 moles of barium bro. Might we need to go from moles to grams and one mole of barium bro. Might the mass is 361. g. Okay, so moles cancel and we're left with g. And once we do this calculation we get 4.84g. So the mass for barium bro, might Is equal to 4.84 g. And that is our final answer for the mass of each compound before decomposition. That's it for this problem, I hope this was helpful.

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