The value of Kc for the reaction of acetic acid with ethanol is 3...

Question

The value of Kc for the reaction of acetic acid with ethanol is 3.4 at 25°C:
CH3CO2H1soln2 + CH3CH2OH1soln2 ∆
Acetic acid Ethanol CH3CO2CH2CH31soln2 + H2O1soln2 Kc = 3.4 (a) How many moles of ethyl acetate are present in an equi- librium mixture that contains 4.0 mol of acetic acid,
6.0 mol of ethanol, and 12.0 mol of water at 25 °C?

Accepted Answer

Hello everyone in this video. I calculate for the numbers of metal beauty no eight present in equilibrium. So this occasion right over here that were given some of the moles of each of our regions here. So this first of all go ahead and recognize what information that were specifically given to solve the problem. So we're given the equilibrium constant is equal to nine. We're giving the number of moles Of water, so that's H20. That's equal to four moles. Were also given the number of moles of methanol, the haze, a formula of ch 30. H. So that's at two moles. And last year he also given the numbers of moles of retinoic acid and that has a formula of CH three ch two, ch two, C 02 H. And that's again at two moles. Alright, so let's go ahead and perform our calculations. So for the equilibrium constant equation and we know that that is equal to the concentration of our products, to its racist power of its coefficients. All over to the concentration of our reactant, raise the power of its coefficients. So applying this to our chemical equilibrium equation here, our products is going to be C H three, CH two, CH two, C. 02 and C H three. And the race to power one. So we won't write the one What supplied by the concentration of our H 20. And then for our react inside we have our ch 30. H multiplied by the concentration of ch three, CH two, CH two, C. 02 and H. Alright. So in this equation our calculation here, we're gonna go ahead and let X equal to what we're trying to solve for SAr CH three or the concentration rather of CH three, CH two, Ch two, C 02, C H three. So the number of moles of metal baton, oh it can be obtained this way because the units of volume will cancel out. Alright, again, so following with this four million here or maybe rather the one right below it says we actually have the products and start materials used. Alright, So we know the equilibrium considerate E that's given to us, that's equal to nine for the first one here, we'll just leave it as an X. Because that's our unknown, we're gonna go ahead and multiply this by four moles. Then for I react inside we have two moles of this and then we have another two moles of this, so we can see here for our denominator, this is equal to 42 times two is four. This will automatically cancel with the numerator on top. So we just automatically canceled this like, so so we get that our X. Which we already let equal to the concentration of the metal baton, wait equal to nine. And to be more exact, we can go ahead, put 9.0 moles of metal 10 08. So this right here is going to be my final answer for this problem

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