Calculate the mass of ethyl acetate formed at 25 C from 1.66 moles of acetic acid and 2.17 moles of ethyl alcohol ? . Kc = 4

Question

Luna Chen · Accepted Answer

The theoretical yield is 110 g of ethyl acetate.

To determine the molarities and perform the equilibrium calculation, we require the volumes of the reactants. Volumes (a) Acid acetic #V_a = 1.66 × (60.05 × color(red)(cancel(black)("g")))/(1 × color(red)(cancel(black)("mol"))) × "1 mL"/(1.0497 × color(red)(cancel(color(black)("g")))) = "95.0 mL"# (b) Alcohol #V_e = 2.17 × (46.07 × color(red)(cancel(color(black)("g")))/(1 × color(red)(cancel(color(black)("mol"))) × "1 mL"/(0.789 × color(red)(cancel(color(black)("g")))) = "126.7 mL"# (c) The entire volume V_a + V_e = "95.0 mL + 126.7 mL" = "221.7 mL"#V_"tot" first priorities The figures #color(white)(mmmmmm)"acetic acid" + "ethanol" ⇌ "ethyl acetate" + "water"# #color(white)(mmmmmmmm)"A"color(white)(mml) +color(white)(ml) "B"color(white)(mll) ⇌color(white)(mml) "C"color(white)(mmm) + color(white)(m)"D"# #"I/mol·L"^"-1": color(white)(mm)7.49color(white)(mmmm) 9.79color(white)(mmmmmll) 0color(white)(mmmmmll) 0# #"C/mol·L"^"-1":color(white)(mml)"-"x color(white)(mmmmm)"-"x color(white)(mmmmm)+xcolor(white)(mmmm) +x# #"E/mol·L"^"-1":color(white)(mm) 7.49-xcolor(white)(m) 9.79-xcolor(white)(mmmm) x color(white)(mmmmmll)x# #K_"c" = (["C"]["D"])/(["A"]["B"]) = (x·x)/((7.49-x)(9.79-x)) = 4# #x^2 = 4(7.49-x)(9.79-x) = 4(73.33 - 17.28x + x^2) = 293.3 - 69.12x + 4x^2# #3x^2 - 293.3 + 69.12x = 0# #x equals 5.61# 5.61 mol/L is the equilibrium concentration of ethyl acetate. When there is little to no change in volume, the mass of ethyl acetate produced is #0.2217 color(red)(cancel(black)("L"))) × (5.61 color(red)(cancel(black)("mol"))))/(1 color(red)(cancel(black)("L")))) × "88.10 g"/(1 color(red)(cancel(black)("mol")))) = "110 g"# 110 g is the ethyl acetate theoretical yield.

Cooper Anderson · Answer

undefined 214.4 grams.