[verun]

Hello:

I just had a quick chemistry question:
Calculate the volume of NH3 produced when 20.0 grams of nitrogen and 3.50 grams of hydrogen react if the pressure equals 1.25 atmosphere and the temperature is 25°C in the production of ammonia.
Balanced Equation: 3H2(g)+N2(g)=2NH3(g)

This is how I figured it out:
20.0g N2(g) * 1.00 mol. N2/28.0 g N2=.714 mol. N2
3.50g H2(g) * 1.00 mol. H2/2.02 g H2=1.73 mol. H2

.714 mol. N2 * 2.00 mol. NH3/1.00 mol. N2=1.43 mol. NH3
1.73 mol. H2 * 2.00 mol. NH3/3.00 mol. H2=1.15 mol. NH3

I believe that H2(g) is the limiting reagent here so I did the following formula using 1.15 mol. NH3 produced.
V=nRT/P

V=(1.15 mol. NH3)(.08206 L*atm/K*mol)(298K)/1.25 atm)
I found V to be 22.5 L.

Am I correct???

[pear]

The calculations look correct.  If you pulled the correct data from tables and calculated correctly, should be right!