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

[Borek]

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?

 

[matt222]

no it doesn't change!

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

 

[Borek]

Yes.

 

[matt222]

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,

 

[Borek]

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.

 

[matt222]

so how to get the change of n now?

 

[Borek]

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.

 

[matt222]

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