Answer to Question #137774 in Chemistry for Shania

Question

C6 H12 O6 + 2C2 H5 OH+ 2CO2
if 3.44 mol of ethanol are produced during the reaction, how many glucose have been used?

Answer 1

Solution:

The chemical equation for alcoholic fermentation is:

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

According to the equation: n(C₆H₁₂O₆) = n(C₂H₅OH)/2

Hence,

n(C₆H₁₂O₆) = 3.44 mol / 2 = 1.72 mol

The molar mass of glucose (C₆H₁₂O₆) is 180.156 g/mol.

m(C₆H₁₂O₆) = n(C₆H₁₂O₆) × M(C₆H₁₂O₆) = 1.72 mol × 180.156 g/mol = 309.87 g

1.72 mol of glucose corresponds to 309.87 g of glucose.

Answer: 1.72 mol of glucose have been used that corresponds to 309.87 g of glucose.