So I do know how to work out letter a) but I'm still very confused with parts b and c. Appreciate the help!

2FeN(s) + 3H2(g) --> 2Fe(s) + 2NH3(g)

Suppose that the reaction is carried out at 300C and 3.00 atm. 1.00 grams of iron (III) nitride reacts.

a)What is the volume of ammonia formed under these conditions if iron (III) nitride reacts completly?

=0.224ml

b)What volume of molecular hydrogen is required?

c) Ammonia condenses to form a liquid at -33.35C at atmospheric pressure. The density of liquid ammonia at this temperature is 0.6818g/ml. If the ammonia from the previous problem is condensed at these temperatures, what woul be its volume?

I think that is 0.224 L (not mL) for a.

Isn't b worked just like a? I assume conditions are 300 C and 3 atm?
moles FeN = 1.00/69.85 = ?
moles H2 required =
moles FeN x (3 moles H2/2 moles FeN) = ?
Then use PV = nRT to solve for V of H2.

c. You have 0.224 L NH3 at 300 C and 3.00 atm.
How many moles was that? 0.01432 (I know that's too many places but you can round off at the end). Convert to grams.
g NH3 = moles NH3 x molar mass NH3.
Then mass = volume x density. You have density and mass, solve for volume at this T of -33.35 C.
The secret to part c is to keep all that stuff about 300 C and volume out of your mind and look at moles. It doesn't matter what T and P are, n is n is n is n. So you go back to basics, use n to solve for grams, then use density to convert to volume of the liquid.