Most wine is prepared by the fermentation of the glucose in grape juice by yeast: c6h12o6(aq) --> 2c2h5oh(aq) + 2co2(g) how many grams of glucose should there be in grape juice to produce 725 mls of wine that is 11.0% ethyl alcohol, c2h5oh (d=0.789 g/cm3), by volume?

123.41 g

Explanation:

Given that the ethyl alcohol produced is 11.0 % by volume.

It means that 1000 mL contains 110 mL of ethyl alcohol

Given that the volume is:- 725 mL

So,

Volume of ethyl alcohol = = 79.75 mL

Given that:- Density = 0.789 g/cm³ = 0.789 g/mL

So, Mass = Density*Volume = = 62.92 g

Calculation of the moles of ethyl alcohol as:-

Molar mass of ethyl alcohol = 46.07 g/mol

The formula for the calculation of moles is shown below:

Thus,

According to the reaction:-

2 moles of ethyl alcohol is produced when 1 mole of glucose reacts

Also,

1.37 moles of ethyl alcohol is produced when mole of glucose reacts

Moles of glucose = 0.685 Moles

Molar mass of glucose = 180.156 g/mol

Mass = Moles*Molar mass = = 123.41 g