Question

How much Hydrogen can be extracted from a gallon of water? Is there any reason why a fuel cell powered aquatic vehicle (boat, submarine) could not supplement the Hydrogen supplies on-board with Hydrogen extracted from sea water, thus extending their range?

Asked by: Michael Ogilvie

Answer

The amount of hydrogen extracted from a gallon of water can be found very easily using the molecular weight of H20 (water), Hydrogen and Oxygen, along with mass conservation. the molecular weight of water is 2 H (molecular weight 1) + 1 Oxygen (Molecular weight 16) for a total of 18. And for every Molecule of water converted, we would get 2 Molecules of Hydrogen.

So, now the question is, how many molecules of water are there in a gallon of water? The density of water is 1g/(cm3) so in 1 gallon of water ( about 3.785 Liters or 3785 cm3) the mass of the water is, 3785g. 1 mole of 6.02x1023 molecules of water is equal has the mass in grams equal to the molecular weight or 18 grams per mole. so 3785 grams corresponds to about 1.265 x 1026 molecules of water.

Now, if every single one of those molecules were converted into Hydrogen we would get twice as much hydrogen as we had of water. or 2.53 x 1026 molecules of hydrogen. however since hydrogen is a diatomic molecule, meaning that the hydrogen that we talk about is H2, we would get 1.265 x 1026 molecules of hydrogen. at 1 atmospheric pressure and 273K, 1 mole of hydrogen fills approximately 22.4L of volume. so 1.265 x 1026 molecules or about 210 moles, would fill 4707 Liters of volume.

As for the reason that you cannot merely extract hydrogen to continuously power your aquatic vehicle, it is due to simple conservation of energy. Simply put, the energy you use to convert the water in the ocean into hydrogen would be at most equal to the energy that hydrogen could provide. Of course due to frictional losses, from things like resistance in your apparatus, you would actually end up using up more energy to get the hydrogen than you would gain from using it as a fuel source.

Answered by: David Seo, Physics Undergraduate Student, UCB, Berkeley

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