GChem volume problem

[113zami]

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Given, 3H2(g)+ N2(g) ---> 2NH3(g)

if you start with 5L of H2(g) and 4L of N2(g) at STP, what will the volume of the threee gases be when the reaction is compolete?

this is how they solved it,

5L of H2(1L of N2/3L of H2)= 1.7 Lof N2 consumed

amount of N2 remaining is 4-1.7= 2.3L

volume of NH3 produced: 5Lof H2(2L of NH3/3L of H2)= 3.3L of NH3

my question is not about how they came up with this answer per se, I just don't understand why did they assume that 3moles of H2 is the same as 3L of H2 ??

I read their explanation and i know that 1mole of any gas occupies 22.4 L at STP but I still don't see why 3mol=3L ?

thanks in advance

 

[bodiddely]

Since both gases are at STP, the number of moles will vary as the volumes; thus the ratio of liters can be used as the ratio of the moles. It's not saying 1mol=1liter, but rather 1 liter of H2:number of moles of H2 as 1L of NH3:number of moles of NH3.

 

[113zami]

ahha..thanks a bunch i knew it had something to do with ratio/proportion stuff just couldn't put my hands on it