A 'canned heat' product used to warm buffet dishes consists of a homogeneous mixture of ethanol 1C2H5OH2 and paraffin, which has an average formula of C24H50. What mass of C2H5OH should be added to 620 kg of the paraffin to produce 8 torr of ethanol vapor pressure at 35 °C? The vapor pressure of pure ethanol at 35 °C is 100 torr.

Video transcript

Hello everyone today. We are being given the following problem and asked to solve for it. So it has a homogeneous mixture of ethyl alcohol and calcium acetate is used to produce heat. If 15 tour of ethyl alcohol vapor pressure is needed, how much alcohol in grams should be mixed with 19 g of calcium acetate at 30 degrees Celsius. Everything goes on to tell us that the vapor pressure of pure ethyl alcohol is 92 tour at degrees Celsius. So the first thing we must do is we must find the mole fraction of our ethyl alcohol which can be abbreviated as C two H five oh H. And so we're going to denote that with X. And so to find that we're going to say that the partial pressure of this C two H five oh H. Is going to be placed over the partial pressure of that same Where the total pressure of that compound. And so when we do this, we say that we have 15/92, which gives us a mole fraction of .1630. And so what we're gonna do now is we are going to find the moles. So we're gonna say we're gonna find the moles of our calcium acetate. We're gonna abbreviate that as such. So to find out we're going to take the mass of calcium acetate that we have. We're going to convert that to grams and we can do that by using the conversion factor that one kg is equal to 10 to the third grams. And then additionally, We're going to take that and we're going to divide that by our total molar mass of that calcium acetate which is 150 58.17 g per mole. And in doing so, we end up with a final answer of 120.1239 moles of our calcium acetate which is C A C H three C. O. And so if we were to finally we have to let y let the variable Y equal the moles of our calcium acetate. So C A C H three C O minus. Then we can see that the mole fraction of C two H five O. H. Is equal to 0.1630. We can also then say this is equal to y over our Y. Plus our moles of calcium acetate which is 20.1239. Rearranging this equation, We get that Y is equal to 0.1630, multiplied by Y. Plus that 120.1239. Further solving further solving this equation, we would get 0.837 Y is equal to 19.580. And we finally get that Y is equal to 23.39. Therefore we can say that we have 23.39 moles of RC 2 H50 H. And and to do that and to convert that into our kilograms which is what we need to do. We need to multiply this first by the molar mass of that compound so we can see that one mole of that C two H five oh H is equal to 46 g. And then to get rid of grams to get our kilograms, we can use the conversion factor that one kg is equal to 10 to the third regular grams. And when we do that, our units cancel and we end up with an answer of 1.2 kg. I hope this helped. And until next time.