Aluminium + water = Hydrogen, QUESTION

[Hydro-Cell]

Hopefully someone can help with this question.

if you were to put 1kg of pure aluminium in water, how much hydrogen could potentially be released from the water???

please ignore the fact that aluminium does not normally react with water.

this is purely figures to do with the atomics i would guess. i did see someone calculate how much hydrogen was contained in 1 litre of water i guess it would just be a case of finding out how much oxygen 1kg of aluminium can consume and multiply it by 2 for amount of hydrogen released.

then all that has to be done is work out how many hydrogen atoms there are to 1 litre of gas and we should be close to a useable figure.

well thats the theory anyway, hope you can help me

[fritznien]

its very simple
       2Al + 3H2O  =  Al2O3 + 3H2
or 56 grams of aluminum can make 6 grams of hydrogen
i kg makes  107 grams of hydrogen,
the real problem is the cost of aluminium. have you priced the stuff?

[Hydro-Cell]

i have priced it and i can get it pretty cheap, i work for an aluminium casting factory and they can get pure aluminium for less than scrap value, but then they do buy 20 tonnes a week

[fritznien]

what do you call cheap? and why would anyone sell for less than scrap price?
but most important is will it be cheaper than other easier to use fuels.
i priced scrap aluminium on google a while back it was from 500 to 2000 dollars a ton.
the hydrogen it would produce would have the energy of 350 litres of gasoline.
so unless your getting it really cheap then your better to buy fuel and cash in the scrap. then take the ol lady to Vegas with the profit! :-)

[tinu]

Hi fritznien,

Al and NaOH +H20 was the most common reaction I’ve used to get small amounts of H2 within a self-pressurizing container, as to be used for filling up balloons for kids (as the needs arrived ) and for other garage experiments (H2 torch etc.).

I’m physicist and obviously, as follows from bellow, such a simple chemical reaction is something that I could not grasp. 
It goes like that:
1. At high enough NaOH concentration â€" close to saturation, the reaction exothermically proceeds at a good rate and a black precipitate forms (as far as I know is a kind of Al(OH) form â€"please help with valences- or maybe something more complex?). Temperature is raising fast, possibly up to the boiling point if cooling is not provided.
2. After a while, presumably after the concentration of NaOH has decreased bellow a certain value, the reaction slows down significantly. By this stage, the O2 in the closed container is consumed (I could measure a significant drop in pressure and I don’t see what else could be consumed from the air if not O2; add the fact that the mix does not seem to be explosive hence H2 is good for a torch:). Temperature is not increasing any longer and it starts to drop, yet the reaction is still exothermic (obvious). I couldn’t tell for sure if the reaction comes to a stop when O2 is fully consumed but I’d say it doesn’t stop but it continues at a quite slow rate.
3. If oxygen is added (i.e. the tank is open and ventilated adequately), reaction proceed again similar to 1 or even faster (and that’s despite the fact that some NaOH must have been consumed), but at some point, the product turns into a whitish-gray precipitate instead of the black one. Now the reaction is very exothermic again.
4. If O2 is available, the reaction goes on until one reactant is fully consumed and imho something is acting as a catalyst because NaOH does seem to work at this stage even at very low concentration, as long as Al and O2 is sufficient.

Whitish-gray precipitate I think is Al2O3, right?
Could you please describe what other products are present and what factors influence the outcome?
I couldn’t solve the riddle by searching and reading on the internet.
Maybe you can help shed some light and eventually correct the mistakes I’ve made in my deductions from observing crude experiments.

Many thank for your help,
Tinu