# What molarity of sodium nitrate do you produce from 4 g of nitric acid and 3 g of sodium hydroxide?

Sep 4, 2016

Unknowable with the given information.

#### Explanation:

$H N {O}_{3} \left(a q\right) + N a O H \left(a q\right) \rightarrow N a N {O}_{3} \left(a q\right) + {H}_{2} O \left(l\right)$

We can work out the molar quantity from the given masses of the reactants:

$\text{Moles"_"Nitric acid}$ $=$ $\frac{4 \cdot g}{63 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.063 \cdot m o l$

$\text{Moles"_"NaOH}$ $=$ $\frac{3 \cdot g}{40 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.075 \cdot m o l$

Clearly, nitric acid is the limiting reagent. And thus $0.063 \cdot m o l$ sodium nitrate could be produced, approx. $5.5 \cdot g$.

This question is very artificial inasmuch as it quotes a mass of nitric acid, when it should quote a volume of acid with a given concentration.

$\text{Molarity}$, the which was asked for in the question, $=$ $\text{Moles"/"Volume of solution}$. We have no way of assessing this quantity.