# If the inventory of the reactor of about 20,000kg

• matt222
In summary: I am sorry, I am really confused because I cant understand the concept.As you know that for 2H2O we have 4kmol H2, so for 1H2O we have 2kmol H2 2kmol H2=2x2kg/kmol=4kg so 4kg or 2kmol H2 produce 124m^3 so 2kmol H2 produce 62m^3 is this wrong?Yes. You are missing the fact that you have 20,000kg of Zr, which is not going to be completely reacted with water - that's why you have to use 10% (or whatever percent) in your calculations

Judging from the question there is no air involved, just water steam and hydrogen. But even if there is air present - does amount of air change?

no it doesn't change!

so in this case I should measure the change on n between water steam and hydrogen gas right!

Yes.

we had already measured n for the hydrugen , now as before should I do the balance equation and find n for 2H20, after that add them together to have total n,

matt222 said:
we had already measured n for the hydrugen , now as before should I do the balance equation and find n for 2H20

Yes - and no. Yes, that's the correct approach. No, you don't have to do it. Just looking at the reaction equation coefficients you should see that water/hydrogen ratio is 1 - number of moles of hydrogen produced equals number of moles of water consumed. As you have already calculated number of moles of hydrogen, you don't have to calculate number of moles of water - it will be the same number.

after that add them together to have total n,

Beware about adding - water is consumed, so it is removed from the reactor.

so how to get the change of n now?

I have spoonfeed you for almost three pages, my hand hurts. It is the highest time you start to think on your own. Especially taking into account fact that what you are asking about now has nothing to do with chemistry, common reasoning is enough.

of course I appreciate your help but you will be shocked if I tell you that I am a computer science background and moved to this field just recently that's why I need a time

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