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.
