Question

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

Answer 1

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

Explanation:

Concentrated nitric acid, is approx. a `15*mol*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:

`Na_2CO_3(aq) + 2HNO_3(aq) rarr 2NaNO_3(aq) + H_2O(l) + CO_2(g)uarr`

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

`"Moles of nitric acid"` `=` `1*Lxx15*mol*L^-1=15*mol`.

Given the stoichiometry of the equation, this will neutralize `"HALF an equiv"` of `Na_2CO_3`, and thus `7.5*mol`.

This has an equivalent mass of `7.5*molxx105.99*g*mol^-1=794.9*g`.