Please help?

Molar mass of CH₃CH₂OH (ethanol) = (12×2 + 1×6 + 16) g/mol = 46 g/mol
Molar mass of CH₃COOH (ethanoic acid) = (12×2 + 1×4 + 16×2) g/mol = 60 g/mol

Initial number of moles of CH₃CH₂OH = (8.28 g) / (46 g/mol) = 0.129 mol
Initial number of moles of CH₃COOH = (60 g) / (60 g/mol) = 1 mol

_____________ CH₃CH₂OH(l) + CH₃COOH(l) ⇌ CH₃COOCH₂CH₃(l) + H₂O(l) …… Kc
Initial: ________ 0.129 mol ______ 1 mol ________ 0 mol _______ 0 mol
Change: ________ -y mol _______ -y mol _______ +y mol ______ +y mol
Equilibrium: __ (0.129 - y) mol __ (1 - y) mol ______ y mol _______ y mol

No. of moles of CH₃COOH at equilibrium:
1 - y = 49.74/60
y = 0.171

Number of moles of CH₃CH₂OH at equilibrium = (0.129 - 0.171) mol < 0 mol
This is impossible.
The question is a typo.