Please help. Thanks?

_____________ CH₃CH₂OH(l) + CH₃COOH(l) ⇌ CH₃COOCH₂CH₃(l) + H₂O(l) …… Kc = 4
Initial: __________ 2 mol _______ 2 mol _________ 0 mol _______ 0 mol
Change: ________ -y mol _______ -y mol ________ +y mol ______ +y mol
Equilibrium: ___ (2 - y) mol ____ (2 - y) mol _______ y mol _______ y mol

Assume that the volume of the reaction mixture is V L.
Kc = [CH₃COOCH₂CH₃] [H₂O] / ([CH₃CH₂OH] [CH₃COOH])
4 = (y/V)² / [(2 - y)/V]²
y² / (2 - y)² = 4
y / (2 - y) = 2
y = 4 - 2y
3y = 4
y = 1.33

No. of moles of the ester CH₃COOCH₂CH₃ formed = 1.33 mol
Molar mass of the ester CH₃COOCH₂CH₃ = (12.0×4 + 1.0×8 + 16.0×2) g/mol = 88.0 g/mol
Mass of the ester CH₃COOCH₂CH₃ formed = (1.33 mol) × (88.0 g/mol) = 117 g (to 3 sig. fig.)

2 tonnes