Question

As you may know, ethyl alcohol, C2H5OH, can be produced by the fermentation of grains, which contain glucose, C6H12O6 → +2C2H5OH(l) + 2CO2(g)

a. How many grams of ethyl alcohol are produced from 1.00 kg of glucose?
b. Gasohol, which you might use to fuel your car, is a mixture of 10 mL of ethyl alcohol (density = 0.79 g/mL) per 90 mL of gasoline. How many grams of glucose are required to produce the ethyl alcohol required for one liter of gasohol?

Answer 1

Answer:

a. 510.6 g of C₂H₅OH are produced from 1kg of glucose

b. 171.1 g of glucose are required

Explanation:

Chemist reaction is this:

C₆H₁₂O₆  →  2C₂H₅OH(l) + 2CO₂(g)

So 1 mol of glucose can produce 1 mol of ethyl alcohol.

First of all, we should convert the mass to g, afterwards to moles

1 kg . 1000 g/ 1kg = 1000 g . 1 mol/180 g = 5.55 moles

Then we can think, this rule of three

1 mol of glucose can produce 2 moles of ethyl alcohol

Then 5.55 moles of glucose may produce the double of moles of C₂H₅OH

(5.55 .2)/1 = 11.1 moles.

Let's convert the moles to mass → 11.1 mol . 46g /1mol = 510.6 g

b. Let's determine the liters of ethyl alcohol we need.

1 gasohol is 10 mL C₂H₅OH / 90 ml of gasoline. We should make a rule of three.

In 90 mL of gasoline we have 10 mL of C₂H₅OH

In 1000 mL (1L) we would have (1000 . 10)/ 90 = 111.1 mL

Now we have to determine the mass of C₂H₅OH that is contained in the volume we have calculated. We must use the density.

Density = Mass /Volume

0.79 g/mL = Mass / 111.1 mL

0.79 g/mL . 111.1 mL = 87.7 g

Now, we convert the mass to moles → 87.7 g . 1mol/ 46g = 1.91 mol

Ratio is 2:1 so 2 moles of C₂H₅OH are produced by 1 mol of glucose

Therefore 1.91 mol would be produced by (1.91 .1)/2 = 0.954 moles

Finally we convert the moles of glucose to mass:

0.954 mol . 180 g/ 1mol = 171.7 grams.