Under laboratory conditions of 25.0 degrees C and 99.5 kPa, what is
the maximum number of cubic decimeters of ammonia that could be
produced from 1.50 dm3 of nitrogen according to the following
equation:
        N2 (g) + 3H2 (g) ---> 2NH3 (g)
a. 3.22 dm3
b. 3.00 dm3
c. 2.70 dm3
d. 3.33 dm3

Here again!!

n = P.V / R.T

For the Nitrogen we have:
P = 99.5 kPa = 0.982 atm     (1 kilopascal = 0.0098692 atm)
T = (25+273)K = 298K
V = 1.5dm3 = 1.5 L
R = 0.082058 atm/(mol * K)

n = (0.9819886*1.5)/(0.082058*298) = 0.06024 mol

Each mol of N2 produces 2 moles of NH3, then we will have (2*0.06024)=
0.12048 moles of NH3.
At the laboratory conditions:
P = 0.982 atm     
T = 298K

The volume of the NH3 will be:

V = (n.R.T)/P = ((0.12048*0.082058*298)/0.982) L = 3 L = 3 dm3

The correct answer is b.

Note:
This problem confirm the following reasoning:
At the same condition of pressure and temperature one mol of different
gases will take up the same volume.
We have an unknown amount of n moles of N2 that takes up 1.5 L, this n
moles will produce 2n moles of NH3, which will be at the same
conditions; this means that this 2n moles of NH3 will take up 2*1.5 L
= 3 L.


Hope this helps.

Regards.
livioflores-ga