A yield of #NH_3# of approximately 98% can be obtained at 200 deg C and 1,000 atmospheres of pressure. How many grams of #N_2# must react to form 1.7 grams of ammonia, #NH_3#?

1 Answer
May 26, 2017

Answer:

1.429 g of #N_2#

Explanation:

We always have to read the problems completely to understand the question. In this case even when we have pressure and temperature conditions specified, they don't matter, first because they are constant and in second because the question ask for grams of #N_2# to produce grams of #NH_3#.

We cannot compare grams of a substance to another in chemical reactions, we need to use moles, then we have to convert the weight of NH3 to moles.

#eta = m//MW#
#MW_(NH_3) = 14.007 + 3(1.008) = 17.031 g//mol#
:. #eta = 1.7g//17.031 g//mol = 0.100 mol#

The yield is 98%, then the number of moles that would be produced are:

#0.100mol//0.98 = 0.102 mol# of #NH_3#

From the reaction (the hydrogen was implicit in the problem, because you are obtaining #NH_3#):

#N2 + 3H2 rarr 2NH3#

There are 2 parts of #NH_3# that are obtained from 1 part of #N_2#

Then the moles of #N_2# required are going to be the half of the #NH_3# moles produced.

#N_2# moles = #NH_3 "moles"//2 = 0.102mol//2 = 0.051 mol#

Last step is too convert the moles back to grams using the MW of #N_2 #

#m_(N_2) = mol xx MW = 0.051mol xx (14.007g//molxx2) = 1.429 g#