# 1.Ethanol (`C_2H_5OH` ) and acetic acid (`CH_3COOH` ) react to form ethyl acetate (`CH_3COOC_2H_5 ` ) according to the following reaction:   `C_2H_5OH (aq) + CH_3COOH (aq) -> CH_3COOC_2H_5 (aq) + H_2O (l)` 125 mL of a 2.00 M ethanol solution is combined with 125 mL of 1.0 M acetic acid solution and the solution is warmed to 98°C. How many grams of ethyl acetate can we expect to produce assuming the reaction goes to completion? First, check if the reaction equation is balanced. Since it is already balanced, we can now proceed to stoichiometric computation to determine the amount of ethyl acetate to be formed in the reaction.

`C_2H_5OH (aq) + CH_3COOH (aq) -> CH3COOC_2H_5 (aq) + H_2O (l)`

Next, we get the amount in...

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First, check if the reaction equation is balanced. Since it is already balanced, we can now proceed to stoichiometric computation to determine the amount of ethyl acetate to be formed in the reaction.

`C_2H_5OH (aq) + CH_3COOH (aq) -> CH3COOC_2H_5 (aq) + H_2O (l)`

Next, we get the amount in moles of both reactants.

`Mol es C_2H_5OH= 2.0 (mol es)/(L) * (125/1000) L`
`Mol esC_2H_5OH = 0.25 mol es`

`Mol esCH_3COOH =1.0 (mol es)/(L) * (125/1000) L`
`Mol esCH_3COOH = 0.125 mol es`

From this we can see that the acetic acid is the limiting reagent since the reaction equation will produce lesser amount of moles of product. Next, we can get the amount of ethyl acetate produced by stoichiometry.

`0.125 mol esCH_3COOH * (1 mol e CH3COOC_2H_5)/(1 mol e CH_3COOH) * (88.105 grams)/(1 mol e CH3COOC_2H_5)`

= 11. 0 grams of ethyl acetate