The reaction of ethylene oxide with water to give ethylene glycol (automobile antifreeze) occurs in 96.0% actual yield. How many grams of ethylene glycol are formed by reaction of 35.0 g of ethylene oxide? (For ethylene oxide, MW = 44.0 amu; for ethylene glycol, MW = 62.0 amu.)

Question

The reaction of ethylene oxide with water to give ethylene glycol (automobile antifreeze) occurs in 96.0% actual yield. How many grams of ethylene glycol are formed by reaction of 35.0 g of ethylene oxide? (For ethylene oxide, MW = 44.0 amu; for ethylene glycol, MW = 62.0 amu.)

Chemical reaction diagram showing ethylene oxide reacting with water to form ethylene glycol, with yield information.

Accepted Answer

Welcome back everyone. The hydrogenation of pure iodine C five H five N produces pie parade in C five H 10 and H. The actual yield of the reaction below is 83%. Calculate the mass of pure Dean and Graham's produced from 42.5 g of protein with a molecular weight for Pepperdine equal to 85.148 A M use. And the molecular weight of pure iodine equal to 79.1 AM use writing our reaction out. We have pure iodine C five H five N reacting with three moles of hydrogen gas to produce our product Pepperdine C five H 10 N H. So we need to calculate our mass of our product produced from our reactant period in so we're calculating are mass of Pepperdine C five H 10 and H which will find by taking our massive paradigm given in the prompt as 42. g of C five H five N period in multiplied by The molar mass of period in which given from our prompt we have in the denominator, a molecular weight of 79.1 AM use. And recall that one a.m. U is equivalent to one g per mole. So we can interpret that as 79.1 g of pure Dyn C five H five N equivalent to one mole of period Ayn in the numerator. So canceling out grams of parodying. We can now go from moles of parodying in the denominator as a multiplier to most of our product Pepperdine in the numerator C five H 10 and H in which from our balanced equation, we see that we have a 1 to 1 molar ratio between our reactant period in and our product Pepperdine. So canceling out moles of period in we can now go from moles of Pepperdine in the denominator as a multiplier, two g of Pepperdine as a multiplier in the numerator. So C five H 10 and H and from our prompt were given the molecular weight of Pepperdine in am you as 85. 48 AM us. And we're going to recognize that this is also understood as 85.1 48 g of protein in the numerator equivalent to one mole of Pepperdine in the denominator. So canceling out moles of Pepperdine, we're left with grams of Pepperdine as our final unit, which is what we want. And this gives us our mass of Pepperdine in our calculators equal to a value of 45.79 or sorry 0.74, 956 g as our massive piper teen. And this is going to be our theoretical yield were given from the prompt, an actual yields of 83%. And we should recall that our formula for percent yield is calculated by taking our actual mass of Pepperdine. So our actual mass of our product, in this case, divided by our mass of our theoretical yield, Which is then multiplied by 100 And so plugging in our values that we know we have our percent yield of 83.0% set equal to our actual mass of Pepperdine. which we are solving for. So our mass of C five H 10 N H in the numerator divided by a mass of our theoretical yield which we just solved above as 45.74956 g of purporting in our denominator. And then this is multiplied by 100. So to isolate for our mass of Pepperdine, our actual massive Pepperdine, we're going to multiply both sides by our denominator 45.74956 g of Pepperdine so that it cancels out on the right and we're going to divide by 100. So we're doing this on both sides to cancel those terms out And sorry, this should be multiply and then divide by 100. And so this will fully isolate our actual mass of Pepperdine C five H 10 N H equal to a value of 37.97 which we can round to two sig figs as 38.0 g of Pepperdine, C five H 10 N H. And this would be our final answer highlighted in yellow as our actual mass of Pepperdine corresponding to choice B in the multiple choice to complete this example. So I hope this made sense and let us know if you have any questions.

The reaction of ethylene oxide with water to give ethylene glycol (automobile antifreeze) occurs in 96.0% actual yield. How many grams of ethylene glycol are formed by reaction of 35.0 g of ethylene oxide? (For ethylene oxide, MW = 44.0 amu; for ethylene glycol, MW = 62.0 amu.)

Verified video answer for a similar problem:

Key Concepts

Stoichiometry

Molar Mass

Percent Yield