Answer to Question #236028 in Chemistry for arissa

Question

Ethyl acetate (CH3CO2C2H5) is used to decaffeinate coffee beans and tea leaves. A solution is prepared by
reacting 10.00 mL of ethanol (C2H5OH) with 10.00 mL of acetic acid (CH3COOH). The densities of acetic
acid and ethanol are 1.0492 g/mL and 0.7893 g/mL, respectively. The equation for the reaction is shown
below:

C2H5OH(l) + CH3COOH(l) → CH3COOC2H5(l) + H2O(l)
(a) Determine the limiting and excess reactant.
(b) Determine the volume (in mL) of the excess reactant when the reaction ends.

Answer 1

C₂H₅OH(l) + CH₃COOH(l) → CH₃COOC₂H₅(l) + H₂O(l)

According to the reaction, n (C₂H₅OH) = n (CH₃COOH).

n = m /M

Density = m / V

m = density x V

m (C₂H₅OH) = 10.00 x 1.0492 = 10.492 g

m (CH₃COOH) = 10.00 x 0.7893 = 7.893 g

M (C₂H₅OH) = 46.07 g/mol

M (CH₃COOH) = 60.052 g/mol

n (C₂H₅OH) = 10.492 / 46.07 = 0.22 mol

n (CH₃COOH) = 7.893 / 60.052 = 0.13 mol

Therefore, CH₃COOH is the limiting reactant. The excess amount of C₂H₅OH is 0.22 - 0.13 = 0.09 mol. This equals to:

m (C₂H₅OH)_excess = n x M = 0.09 x 46.07 = 4.1463 g

V (C₂H₅OH)_excess = m / density = 4.1463 / 1.0492 = 3.95 ml