Question

Hi,
A neutralization reaction was performed using NaOH (volume = 49.0 mL, concentration = 0.600 M ) and nitric acid (volume = 49.0 mL, concentration = 0.750 M ).
Calculate the moles of the limiting reactant.

NaOH + HNO 3 → NaNO 3 + H 2 O

Please Give your answer to three significant figures.Thanks

Answer 1

NaOH + HNO$_{3}$ $\to$ NaNO$_{3}$ + H₂O

Solution:

step 1: find the moles of reactants in terms of molarity and volume

$C_{M}= \tfrac{n_{solute}}{V_{solution}}$ $\to$ $n_{solute}$ = $C_{M} * V_{solution}$

n(NaOH) = 0.0294 moles

n(HNO$_{3}$) = 0.03675 moles

step 2: determine the limiting reactant

given reaction is equal so we should just compare the moles of reactants to each other.

NaOH + HNO$_{3}$ $\to$ NaNO$_{3}$ + H₂O

0.0294 < 0.03675

So moles of sodium hydroxide are smaller than nitric acid. Therefore, limiting reactant in this reaction is sodium hydroxide

limiting reactant $\to$ NaOH

moles of limiting reactant $\to$ 0.0294 moles