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

Question

Jeremiah Adams · Accepted Answer

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

When nitric acid is stoichiometric,

#"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#

Owen Ambrose · Answer

undefined The molar mass of NH₃ is 17.03 g/mol and that of NH₄NO₃ is 80.05 g/mol. Using the stoichiometry of the reaction between NH₃ and HNO₃, where 1 mole of NH₃ reacts with 1 mole of HNO₃ to produce 1 mole of NH₄NO₃, we can calculate the amount of NH₄NO₃ produced.

First, we convert the mass of NH₃ to moles:
$$ \text{moles of NH}_3 = \frac{15.86 \text{ g}}{17.03 \text{ g/mol}} = 0.93 \text{ mol}$$

Since the reaction is 1:1 between NH₃ and NH₄NO₃, the number of moles of NH₄NO₃ produced will be the same as the moles of NH₃ reacted:
$$ \text{moles of NH}_4\text{NO}_3 = 0.93 \text{ mol}$$

Now, we calculate the mass of NH₄NO₃ produced:
$$ \text{mass of NH}_4\text{NO}_3 = (0.93 \text{ mol}) \times (80.05 \text{ g/mol}) = 74.44 \text{ g}$$

So, 74.44 grams of ammonium nitrate will result from the reaction.