We need the volumes of the reactants, so we can calculate the molarities and do the equilibrium calculation.
Volumes
(a) Acetic acid
#V_a = 1.66 color(red)(cancel(color(black)("mol"))) × (60.05 color(red)(cancel(color(black)("g"))))/(1 color(red)(cancel(color(black)("mol")))) × "1 mL"/(1.0497 color(red)(cancel(color(black)("g")))) = "95.0 mL"#
(b) Ethanol
#V_e = 2.17 color(red)(cancel(color(black)("mol"))) × (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) Total volume
#V_"tot" = V_a + V_e = "95.0 mL + 126.7 mL" = "221.7mL"#
Initial concentrations
#["Acetic acid"] = "1.66 mol"/"0.2217 L" = "7.49 mol/L"#
#["Ethanol"] = "2.17 mol"/"0.2217 L" = "9.79 mol/L"#
The calculations
#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 - 69.12x +293.3 = 0#
#x = 5.61#
The equilibrium concentration of ethyl acetate is 5.61 mol/L.
If the volume change is minimal, the mass of ethyl acetate formed is
#0.2217 color(red)(cancel(color(black)("L"))) × (5.61 color(red)(cancel(color(black)("mol"))))/(1 color(red)(cancel(color(black)("L")))) × "88.10 g"/(1 color(red)(cancel(color(black)("mol")))) = "110 g"#
The theoretical yield of ethyl acetate is 110 g.