# What mass of sodium carbonate is required to react with a 1*L volume of nitric acid, whose concentration is 15*mol*L^-1?

##### 1 Answer
Jan 30, 2017

You need to specify the concentration, the molarity of the nitric acid.

#### Explanation:

Concentrated nitric acid, is approx. a $15 \cdot m o l \cdot {L}^{-} 1$ solution. I will assume you are using this stuff (in fact, the concentration would not be issued to A level students, so you will have to adjust your answer appropriately).

We need (i) a stoichiometrically balanced equation:

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

Mass and charge are balanced as required, and (ii), equivalent quantitites of acid:

$\text{Moles of nitric acid}$ $=$ $1 \cdot L \times 15 \cdot m o l \cdot {L}^{-} 1 = 15 \cdot m o l$.

Given the stoichiometry of the equation, this will neutralize $\text{HALF an equiv}$ of $N {a}_{2} C {O}_{3}$, and thus $7.5 \cdot m o l$.

This has an equivalent mass of $7.5 \cdot m o l \times 105.99 \cdot g \cdot m o {l}^{-} 1 = 794.9 \cdot g$.