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Fri 21 May, 2004 05:21 am
I've got a home-brew kit on the go and was sitting looking at the CO2 bubbling out of the valve at the top and wondered how to calculate the volume of CO2 produced for the fermentation of a certain volume of alcohol. I studied chemistry at A-level (16-18) and know that it is possible to do this, but that was a long time ago (10yrs) and I've forgotten how.
Can anyone help?
There will be 5000ml of rum at the end, with an ABV of around 20%, which equals 1000ml of pure ethanol. Also assume that the CO2 will be at at atmospheric pressure and, say, 20degC for calculating volume.
Thanks!
I'll start at the beginning 1000ml of ethanol has a mass of 800g, this is 17.4 moles.
Sugar breaks down into alcohol and CO2 with the following formula
C6H12O6 -----> 2 C2H5OH + 2 CO2
So for each mole of alcohol produced there will be one mole of CO2 given off.
At room temperature a mole of gas ocupies 24 L so the volume produced will be....
408L
Hope this helps.
Many thanks, Stephen. I was wondering if you could explain how you arrived at 800g for the ethanol (using the atomic weights to calculate the mass of the molecule then multiplying up?)
Do I remember correctly that a mole = 1x10^24 particles?
Sorry should have made that point more clearly....
Ethanol has a density of 0.8 g/ml so 1000 ml of alcohol has a mass of 800 g
The density has to be looked up, I used a chemical supply catalogue, but I'm sure you can find such information on the net.
Ethanol has a molecular weight of 46 g/mol
so 800 g dived by 46 g/mol is 17.4 moles.
A mole is 6.024...... x 10^24 particles
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