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

Mar 7, 2017

$N {H}_{3} \left(a q\right) + H N {O}_{3} \left(a q\right) \rightarrow N {H}_{4} N {O}_{3} \left(a q\right)$

#### Explanation:

Given stoichiometric nitric acid,

$\text{moles of ammonia "-=" moles of ammonium nitrate}$

$\text{moles of ammonia "-=}$ $\frac{15.86 \cdot g}{17.03 \cdot g \cdot m o {l}^{-} 1} = 0.931 \cdot m o l$

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

$0.931 \cdot m o l \times 80.05 \cdot g \cdot m o {l}^{-} 1 \cong 75 \cdot g$