Question

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

Answer 1

`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`