355 KWh corresponds to how many litres of Hydrogen & Oxygen when Electrolysis iof Water is carried out.?
Ok, to solve this problem, you have to know how much energy is exacted from the water formation reaction. It's 572 KJ/mole of water. Since, 1 KWh = 3,600,000 J = 3,600 KJ, 355 KWh corresponds to 1,278,000 KJ, which equals 2234.27 mole water.
2 H2 + O2 ==> H2O
according to the equation, there are 4468.54 mole of H2 and 2234.27 mole of O2.
Use equation to find mass of each substance :
m = n x M.
Mass of H2 : (M= 1)
m = 4468.54 x 2 = 8937.08 g.
The same to mass of oxygen (M = 16), we get 71496.64 g.
We use density equation to find the volume of H2 and O2.
D of H2 = 0.08988 g/L ~ 0.09 g/L
D=m/V ==> V = 8937.08/0.09 ~ 993000.89 (litters)
Do the same to O2 (D= 1.429 g/L) , we get :
V = 50032.63 litters.
I hope I do right on the calculation, but I think the method is right.
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