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

1 Answer
Sep 4, 2016

Answer:

Unknowable with the given information.

Explanation:

#HNO_3(aq) + NaOH(aq) rarr NaNO_3(aq) + H_2O(l)#

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

#"Moles"_"Nitric acid"# #=# #(4*g)/(63*g*mol^-1)# #=# #0.063*mol#

#"Moles"_"NaOH"# #=# #(3*g)/(40*g*mol^-1)# #=# #0.075*mol#

Clearly, nitric acid is the limiting reagent. And thus #0.063*mol# sodium nitrate could be produced, approx. #5.5*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.

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