hello everyone?

Question

Fermentation is a complex chemical process of wine in which glucose in fruit is converted into ethanol and co2
C6 H12 O6 --->2 C2 H5 OH (I) + 2 C02 (g)
Startin with 500.4g of glucose, what is the maximum amount of ethanol (in grams and liters) that can be obatained by the process? (Density of ethanol =0.789g/mL

Answer 1

You have grams, so start with grams. Liters will come later.

First, compute how many MOLES of glucose you have by adding up the atomic weights of each atom. You have 6 carbons at (approximately) 12 g/MOLE each, then 12 hydrogens and 6 oxygens. Add up that number, which gives you the molecular weight (in grams) of one MOLE of glucose. Divide the molecular weight of glucose into the actual amount of glucose you have (in grams), and that number is the number of MOLES of glucose.

OK, now compute from that how many MOLES of ethanol you can get, noting that you get 2 ethanols for every glucose. You are still in MOLES so you aren't home free yet.

Add up the molecular weights of the atoms in ethanol to give you the number of grams in one MOLE of ethanol. Multiply that by the number of moles and you have grams of ethanol. Then, since you have the number of grams and the density, you can do the math to compute the volume in liters.

Note that I don't do someone else's homework, so I didn't compute it for you. But I told you how to compute it. Meet me halfway and do the math.

Answer 2

molecular weights:

C6H12O6 = 180
C2H5OH = 46
CO2 = 44

No. of moles
C6H12O6 = 500.4 g/180 = 2.78

No of C2H5OH moles formed:
= 2*2.78
= 5.56 moles

weight of C2H5OH formed:
= 5.56*46 = 255.76 grams

volume:

= 255.76/0.789 = 324.16 mL