C2H5CO2H(l) + C3H7OH(l) ---> C2H5COOC3H7(l) + H2O(l)(adsbygoogle = window.adsbygoogle || []).push({});

18.5 mL propanoic acid are mixed with 19.0 mL n-propanol to produce n-propyl propanoate when heated and distilled.

At room temperature, the equilibrium constant, Keq is approximately 3. Calculate the number of moles of reactants and products expected in the equilibrium mixture of this reaction. (Hint: Let X be the number of moles of each product formed at equilibrium. Substitute into the equilibrium equation using the moles of each reactant and product at equilibrium each divided by the total reaction volume. Leave the equilibrium concentrations as fractions. You can then take the square root of both sides of the equation after canceling like terms, i.e. the volumes.)

Other Data:

Boiling Point of n-propyl propanoate = 122 degrees C

Volume of water collected = 3.31 mL

Weight of product (C2H5COOC3H7) = 7.4846 g

I have no idea of how to start this problem, can anybody give me a push in the right direction please?

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# Homework Help: Equilibrium in reaction of C2H5CO2H(l) + C3H7OH(l)

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