Question

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`?

Answer 1

`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`