Reaction of gaseous fluorine with compound X yields a sin- gle pr...

Question

Reaction of gaseous fluorine with compound X yields a sin- gle product Y, whose mass percent composition is 61.7% F and 38.3% Cl. (d) How much heat in kilojoules is released or absorbed in the reaction of 25.0 g of X with a stoichiometric amount of F2, assuming 87.5% yield for the reaction?

Accepted Answer

Hi everyone. This problem reads the reaction of two moles of compound A with one mole of oxygen gas produces two moles of a single compound B. The mass percentages of carbon and oxygen and compound B. R 27.3% and 72.7% respectively. The heat of formation of compound B is negative, 393.5 to kill a jules per mole. How much heat and kill a jewel units is released when 12. g of compound A reacts with the stoke eo metric amount of oxygen gas to produce compound be assumed that the reaction yield is 74%. Okay, so for this problem, the question that we want to answer is how much heat and kill a jules is released and were given the grams of compound A. Okay, so here's how we're going to solve this problem. We're going to use the percentage is to find the empirical formula for compound beat. So the first thing that we're going to do is we're going to assume that we have 100 g of compound A By assuming that we have 100 g of compound a. We can take the percentages that were given. So we have carbon is equal to 27.3% and we have oxygen is equal to 72.7%. So we're going to use the percentages to find the empirical formula. So this 27.3% is going to become 27.3 g and the 72.7% is going to become 72.7 g. So we're going to divide the masses by their atomic masses. So let's go ahead and write down the atomic masses. Okay, so the atomic mass for carbon Is 12.01 and the atomic mass for oxygen is 16. So let's go ahead and divide the masses by their atomic masses. So for carbon, our atomic mass is 12.01. And for oxygen, our atomic mass is 16. Okay, So what that becomes then is we get 2. carbon And 4.543754 oxygen. So now we're going to divide both numbers by the smaller number and round off the answer to the nearest whole number. So out of these two we see that the smallest, the smaller number is going to be the number that we got for carbon. Okay, so we're going to rewrite this. So we're going to get Carbon is going to equal 2.27 And oxygen is 4.54375. So now we're gonna divide what we got this is what we got over here for both. Now we're gonna divide by the smaller number and the smaller number is 2.27311. So we're going to divide both of those by the smaller number. And so the number that we're going to get for carbon, we're going to get one and for oxygen, we're going to get 1.9989, which we can say is too alright, so the empirical formula of compound B is what we're finding right now. So we can go ahead and write the empirical formula of compound B is C. 02. Okay, now that we know the empirical formula, since we don't have the molar mass of B, we will assume that the empirical formula is the molecular formula. So we're going to also assume that the molecular formula is C. 02. Okay, so right here, also molecular formula and the reason why we're assuming is because we don't have the molar mass of B. Alright, so now let's go ahead and write a reaction. Okay, so based off of the information in the problem, the reaction for this is going to be too CO2, plus O two yields to C +02. So that is going to be our reaction. The heat released in the reaction be can be determined in this way. So we're going to write this out. So we're gonna take the mass of carbon monoxide. Okay? And we're gonna go from massive carbon monoxide, two moles of carbon monoxide. Then we're gonna use the mole ratio. Okay? And we're going to use the mole ratio between carbon monoxide and carbon dioxide. Okay, so we're using the mole ratio we're not going to go to moles of C. 02. Okay, and from there, we're gonna use the percent yield. So, and that was given in the problem. So, the percent yield to find the actual C. 0. 2 moles. And then from there we're going to use the heat of formation and that is going to give us the heat released. So, this is our roadmap of what we're going to do to determine the heat released in the reaction. Alright, so, first of all, let's go ahead and write out the molar mass for carbon monoxide. So, the molar mass for carbon monoxide for carbon, we have 12.1 plus 16 is equal to 28.01 g per mole. Okay, so, and our molar mass for carbon dioxide Is going to equal 12.01 plus two times 16. So, our moller mask for carbon dioxide is 44.1 g per mole. So, let's go ahead and use our road map to find our heat release. So, we're going to start off with the mass that was given in the problem. We're told that we have 12.6 g of compound A So, that's going to be our starting point. All right. So, we have 12.6 g. All right, And we want to go from grams of carbon monoxide two moles of carbon monoxide. So, in one mole of carbon monoxide, there is 28. g of carbon monoxide. Alright, So, now we want to use our multiple ratio between carbon dioxide and carbon monoxide and we'll do that using the equation that we wrote. So, looking up up above, we're looking at are multiple ratio between carbon monoxide and carbon dioxide. So we see for every two moles of carbon monoxide Consume two modes of carbon dioxide is produced. So let's go ahead and write that here. So, for every two moles of carbon dioxide, There is two moles of carbon monoxide. Alright, so now we want to go, we're gonna use the theoretical or the percent yield to find the actual Carbon dioxide moles. Okay, so, using the theoretical yield, we have 74 moles of carbon dioxide is the actual Over 100 moles of carbon dioxide is a theoretical. Okay, so we'll write actual over theoretical because we're using the percent yield. So this is actual over theoretical. Alright, and now we're gonna use the heat the heat of formation for the compound. Okay, So, we're told in the problem that it's negative 393.52 kg joules per mole. So, we're gonna go ahead and write 393. to kill a jules of heat for every one mole of carbon dioxide. So, let's go ahead and just make sure our units cancel here are grams of carbon monoxide cancel moles of carbon monoxide cancel moles of carbon dioxide cancel moles of carbon dioxide cancel. So here we see that the unit that we're left with is kill a jewels of heat, which is the question that we want to answer here. So once we do this calculation, We'll plug this value, we'll do this calculation in our calculator. And once we do the calculation, we're going to get the heat released is equal to 131 kg jewels. And this is going to be the final answer. Okay, so this is the heat and kill a jewel units released when 12.6 g of compound A reacts with the stoke e. O, metric amount of 02 to produce compound B. That is it for this problem. I hope this was helpful.

Reaction of gaseous fluorine with compound X yields a sin- gle product Y, whose mass percent composition is 61.7% F and 38.3% Cl. (d) How much heat in kilojoules is released or absorbed in the reaction of 25.0 g of X with a stoichiometric amount of F2, assuming 87.5% yield for the reaction?

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