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

Question

Henry Antrim · Accepted Answer

Unknowable with the given information.

#HNO_3(aq) + NaOH(aq) rarr NaNO_3(aq) + H_2O(l)# With the mass of the reactants known, we can calculate the molar quantity: #"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 extremely fabricated because it quotes a volume of acid with a specific concentration rather than a mass of nitric acid. #"Molarity"#, the which was asked for in the question, #=# #"Moles"/"Volume of solution"#. We have no way of assessing this quantity.

Cooper Anderson · Answer

undefined 1. Calculate moles of nitric acid (HNO3) using its molar mass (63 g/mol).
2. Calculate moles of sodium hydroxide (NaOH) using its molar mass (40 g/mol).
3. Determine the limiting reactant by comparing the moles obtained in steps 1 and 2.
4. Use the limiting reactant to find moles of sodium nitrate (NaNO3) produced.
5. Calculate molarity by dividing moles of NaNO3 by the volume in liters.