How many grams of ethanol c2h5oh can be made by the fermentation of 1150 g of glucose?

Answer is: 588.15 grams of ethanol.
Chemical reaction: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂.
m(C₆H₁₂O₆) = 1150 g.
n(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 1150 g ÷ 180.16 g/mol.
n(C₆H₁₂O₆) = 6.38 mol.
From chemical reaction: n(C₆H₁₂O₆) : n(C₂H₅OH) = 1 : 2.
n(C₂H₅OH) = 12.76 mol.
m(C₂H₅OH) = 12.76 mol · 46.07 g/mol.
m(C₂H₅OH) = 588.15 g.

Idon't see the pic?

Qand k are only the same when the forward rate of the reaction has reached zero.