If #15.86*g# of ammonia are reacted with excess nitric acid, what quantity of ammonium nitrate result?

1 Answer
Mar 7, 2017

Answer:

#NH_3(aq) + HNO_3(aq) rarr NH_4NO_3(aq)#

Explanation:

Given stoichiometric nitric acid,

#"moles of ammonia "-=" moles of ammonium nitrate"#

#"moles of ammonia "-="# #(15.86*g)/(17.03*g*mol^-1)=0.931*mol#

We use this molar quantity to calculate an equivalent mass of #"ammonium nitrate"#, given that there is 1:1 equivalence between ammonia and ammonium nitrate in the stoichiometric equation:

#0.931*molxx80.05*g*mol^-1~=75*g#