Given the equation:#N_2+3H_2 rarr 2NH_3#, what volume of #NH_3# at STP is produced if 25 grams of #N_2# is reacted with an excess of #H_2#?

1 Answer
Jan 27, 2016

Answer:

#V = 40 L#

Explanation:

One thing I think it's interesting to note is that you'll never be able to have that reaction happen in the standard temperature and pressure; you can, however, have the resulting ammonia in the STP.

Anyhow, #N_2# has molar mass # 14*2=28#, so for 25 g we have

#n_(N_2) = m/(MM) = 25/28 mol#

Every one mol of nitrogen makes two moles of ammonia, so we have

#n_(NH_3) = 2n_(N_2) = 2*25/28 = 25/14 mol#

Since we know that at STP 1 mol of gas has 22.4 L (this can easily be checked by plugging the appropriate values in #pV = nRT#) We have

#V_(NH_3) = 22.4*25/14 = 40 L#