When ethane 1C2H62 reacts with chlorine 1Cl22, the main product is C2H5Cl, but other products containing Cl, such as C2H4Cl2, are also obtained in small quantities. The formation of these other products reduces the yield of C2H5Cl. (a) Calculate the theoretical yield of C2H5Cl when 125 g of C2H6 reacts with 255 g of Cl2, assuming that C2H6 and Cl2 react only to form C2H2Cl and HCl. (b) Calculate the percent yield of C2H5Cl if the reaction produces 206 g of C2H5Cl.

Video transcript

Hello everyone today. We are being given a problem and asked to solve for theoretical yield. The problem reads as when 14.60 g of beauty and 76.9 g of iodine react. What is a theoretical yield. And then we have the chemical reaction here. And luckily for us it's already balanced so we can move on to step one which is first. We need to find the limiting reactant. So what we're gonna do is we're gonna take even each of our given values. So for example, 14.6 g of Beauty, which is C four H eight. And we're going to convert that into moles of C four H 7. i. So we're going to say one mole of beauty Is equal to its molar mass which is 56 point oh 107g. And then we want moles of C four H seven. I So we're going to do the multiple ratio. We're going to say we have per one mole of C four H eight or Beauty. We have one mole of C four H 7. I our units cancel. And we were given 0.2602 moles of C four H seven. I do the same with iodine. We have 76.9 g of iodine which is I two. We're going to do the same procedure. We're going to use the molar mass. We're going to say one more Is equal to .8 g of I two. And then use the multiple ratio once again. So we're gonna say one mole of C four or C four H seven. I Gives us one more of iodine. Our units are going to cancel. Once again, We're gonna be left with 0.303 Moles of C four H 7 i. The limited reaction is simply the reaction that produces the least amount of product. And in this case that is going to be our beauty. So we're gonna debut, thine is our limiting reacted. Our second step is to take the amount of moles made from our limiting reactant. So we're going to take 0.2602 moles of C four H seven I. And we're going to convert that by using the its molar mass. So we're going to say one mole of C four H seven I. Is equal to 182.0 g of C four 87. I. Are you gonna cancel? And we're going to be left with 47.36 g of C4 H seven i. And that is our theoretical yield. I hope this helped. And until next time