Consider the following reaction: 2N2(g) + 5O2(g) + 2H2O(g) --> 4HNO3(g). If a reaction mixture contains 28g of N2, 150g of O2, and 36g of H2O, what is the limiting reactant?

1 Answer
Nov 1, 2015

Answer:

You have the equation:

#N_2(g) + 5O_2(g) + 2H_2O(g) rarr 4HNO_3#

Explanation:

You have been given the masses of each reagent. These are molar quantiies when we divide them by the molecular mass of each reagent:

i.e. #(28 *g *N_2)/(28* g *mol^-1)#; #(150 *g *O_2)/(32* g* mol^-1)#; #(36* g *H_2O)/(36 *g *mol^-1)#.

It doesn't too much arithmetic to realize that the in terms of molar quantities you have #N_2:O_2:H_2O:HNO_3# #=# #1:<5:2#. The reaction mixture is almost stoichiometric (what do I mean?), but there is a slight deficiency of oxygen gas. The limiting reagent is therefore #??#.