Alkohol -opgaver

Nogen som vil forklare, hvordan jeg udregner de følgende opgaver?

1) Procent af convertet glykose
(sammenligning mellem den teroretiske værdi (før eksperimentet) og den eksperimentale værdi.

2) mængden af produceret ethanol.

3) Volumen af ethanol i reaktiones produktet.
Ethanol har en densitet af 0.789 g/mol

4) Indholdet af ethanol i volume %
Lad os sige at den totale volume af reaktiones produktet er 125mL


Her kommer min raport (på engelsk – grundet noget sammenlægning af nogle fag) :


Alcohol fermentation


Ethanol is produced by the fermentation of glucose. The enzymes in yeast break the glucose down into ethanol and carbon dioxide.

The chemical equation:

C6H12O6(aq) > 2 CO2(g) + 2 CH3CH2OH(l)

Before the experiment:

1) The content of glucose in glucose monohydrate C6H12O6 • H2O

M(C6H12O6) = ((6*12.01) + (12*1.008) + (6*16.00)) ˜180.16 g/mole
M(C6H12O6 • H2O) = (180.16+ ((1.008*2)+16.00)) ˜ 198.18 g/mole

M(C6H12O6) / M(C6H12O6 • H2O) = 0.909 * 100 = 90.9 %

(90.9%/100) * 25g. Of C6H12O6 • H2O = 22.73 g. Of C6H12O6

m/M = n

n(C6H12O6) = 22.73 g. / 180.16 g/mole = 0.126 mole

2) The mass loss

C6H12O6(aq) > 2 CO2(g) + 2 CH3CH2OH(l)

n(C6H12O6) = 0.126 mole
n(2 CO2) = (0.126 mole * 2) = 0.252 mole
n(2 CH3CH2OH) = (0.126 mole * 2) = 0.252 mole

m = n*M

M(CO2) = ((12.01 + (2*16.00)) = 44.01g/mole

m(CO2) = 0.252 mole * 44.01 g/mole = 11.09 g. mass loss

3) Mass of ethanol produced by complete fermentation

M(CH3CH2OH) = (12.01+(3*1.008)+12.01+(2*1.008)+16.00+1.008) = 46.07 g/mole

m(CH3CH2OH) = 0.252 mole * 46.07 g/mole = 11.61 g.

4) The purpose of the calcium chloride pipe

The experiment:

Apparatus and chemicals:
250mL conical flask. Stopper with glass pipe and rubber tube. Graduated cylinder. Scale. Calcium chloride. 25 g. of Glucose monohydrate. 10 g. of crumbled Yeast. Saturated Ca(OH)2 solution. 120mL demineralised water.

Starting the fermentation:
We dissolved the glucose in demineralised water and added the crumbled yeast (in the conical flask). After, we provided the flask with a stopper with glass pipe and rubber tube and measured the whole system: 330.58g.

Investigating the gas product:
The solution was placed for some days and the solution should start after half an hour. The gas produced by the fermentation reaction should then bubble through a solution of saturated calcium hydroxide.

The chemical equation for the reaction:

CO2 + Ca(OH)2 > CaCO3 + H2O

This is an acid-base reaction. CO2 is an acid and Ca(OH)2 is a base and together the produce a salt (CaCO3) and a liquid (H2O).

Finishing the experiment:
We reweighed the system after the fermentation was completed. It weighed 317.30g, which is less than before fermentation, so the mass loss is: 330.58g. – 317.30g. = 13.28g.

After the experiment:

1) Percent of converted glucose
(Comparison between theoretical value (before the ex.) and the experimental value.

2) Amount of produced ethanol

3) The volume of ethanol in the reaction product
Ethanol has a density of 0.789 g/mL

4) Content of ethanol in volume%
Assuming the total volume of the reaction product to be 125mL.