During the fermentation of glucose, ethanol 50.2 g was obtained. the product yield is 85%

During the fermentation of glucose, ethanol 50.2 g was obtained. the product yield is 85% calculate the mass of glucose that has undergone fermentation.

Let’s write the equation for the reaction of glucose fermentation:
С6H12O6 = 2C2H5OH + 2CO2
Let’s find the amount of practically obtained ethanol according to the formula n ‘= m / M = 50.2 g / 46 g / mol = 1.1 mol
Then, theoretically, the amount of ethanol is equal to the expression n = n / u = 1.09 / 0.85 = 1.3 mol
Based on the reaction equation, n (C6H12O6) = 0.5n (C2H5OH) = 0.5 * 1.3 mol = 0.65 mol
Find the mass of glucose that has undergone fermentation according to the formula m = n * M = 180g / mol * 0.65mol = 117g
Answer: mass of glucose = 117g