[ozzlomo]

How many liters of NH3 are produced in the following reaction when 3.0L of N2 reacts with 4.0 L of H2? N2(g) + 3H2(g) --> 2NH3(g)

I'm currently thinking of ways to approach this problem. The first way I can think of solving this problem is using the stoichiometric equivalents, but I don't really have a specific idea. Any help would be much appreciated.

Also, if a sample of nitrogen gas at 298 K and 745 torr has a volume of 37.42 L, how many liters will it occupy if the pressure is increased to 894 torr at a constant temperature?

The "constant" temperature was throwing me off at first, but then I realized they were hinting that the temperature does not change (Please correct me if I'm wrong). Using P1*V1/T1 = P2*V2/T2 I calculated 31.1 L. Is my answer correct (including to the number of significant figures)?

[ozzlomo]

I just got the answer using the limiting reagent. Since nitrogen and hydrogen react in a 1 to 3 ratio, The hydrogen will be limiting (in order to react with 3 L of nitrogen, you would need 9 L of hydrogen, but you only have 4 L).

Therefore:

4 L H2 x (2 L NH3 / 3 L H2) = 2.67 L NH3

Correct?

[Borek]

Twice OK, although I would rather round the second answer to 31.2L.