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Due-Department-8666 t1_iyc0ngg wrote

Because we can capture/filter other emissions with catalytic converters but water just escapes our abilities.

Snaps fingers* Shoot, we were so close too.

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CMDR_kamikazze OP t1_iyc1foa wrote

No, it's ain't a joke really. When you will burn 1 kilogram of hydrogen you will get 9 (yes, NINE) liters of water. Good luck capturing that. Imagine all this water in the winter on the highway. Or up in the stratosphere as clouds.

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Due-Department-8666 t1_iyc2bdi wrote

Got a link for the numbers?

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CMDR_kamikazze OP t1_iyc39o7 wrote

Do you need a link to a chemistry textbook?

Assuming we're burning 1000 grams of hydrogen and have complete combustion and a reaction that is 100% ideal (it won't be like that, but pretty close):

2H2+O2->2H20

With that, amount=mass in grams/Mr or Ar. Hence the amount of H2 would be 500 mol. Using a stoichiometric ratio amount of water is 500 mol.

Now mass = amount×Mr or Ar, hence 500×(2+16)=9000g of water from burning 1000g of hydrogen.

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Helios4242 t1_iyc4q7c wrote

Great now tell me how many moles H2O produced in a combustion reaction from jet fuel providing an equal amount of energy :)

Providing one and not the other is deceptive even if accurate.

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CMDR_kamikazze OP t1_iyc5fwc wrote

BTW while with jet fuel it's about twice as much water produced, how about common car fuels, like gasoline and diesel (OMG I hate diesel, ban this thing for good already), could you calculate it? I'm not very good with hydrocarbons.

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Helios4242 t1_iyc6vdb wrote

for simplicity, I'm assuming octane (C8H18). Energy density is 46MJ/kg, so we need 2.6 kg to match the energy in 1kg H2. MW of 114 means we are using 22.8 moles, producing 9 moles of H2O per mole octane. That's 206 moles, sopretry similar. Diesel isn't gonna be much different.

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