The total volume of hydrogen gas needed to fill the Hindenburg was 2.00 x 10^8 L at 1 atm and 25 C. How much energy was release when it burned?
2 H2(g) + O2(g) ---> 2 H2O(l) delta H = -572 kJ
2 H2(g) + O2(g) ---> 2 H2O(l) delta H = -572 kJ
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If you have written the balance eq correctly, the energy released is 570kJ if you burn 2 mols of hydrogen. This problem can be solved by first calculate the mole of the hydrogen in the Hindenburg, then calculating the energy released.
Finding mole: Let's assume ideal gas law:
PV=nRT ==> n=PV/RT ==> n(H2) = (1 atm · 2.00 · 10^8 L) / (0,0820574587 L · atm · K-1 · mol-1 · 298K)
n(H2) = 8178661 mol
now we can find the energy released:
as stated above 2 mol H2 gives off 572 kJ. that means 1 mol gives off 286kJ.
hence 8178661 mol has to give off
286kJ/mol · 8178661 mol = 23.4 · 10^8 kJ
hope this helps for this and future exercises!
Finding mole: Let's assume ideal gas law:
PV=nRT ==> n=PV/RT ==> n(H2) = (1 atm · 2.00 · 10^8 L) / (0,0820574587 L · atm · K-1 · mol-1 · 298K)
n(H2) = 8178661 mol
now we can find the energy released:
as stated above 2 mol H2 gives off 572 kJ. that means 1 mol gives off 286kJ.
hence 8178661 mol has to give off
286kJ/mol · 8178661 mol = 23.4 · 10^8 kJ
hope this helps for this and future exercises!