# Determine the normality of sodium nitrate (NaNO3) solution formed when an additional 2 L of water was added to the reaction involving 200 mL of 0.35 M barium nitrate (Ba(NO3)2) and 300 mL of 0.2 M sodium sulfate (Na2SO4) as shown below.?

May 7, 2017

The ${\text{NaNO}}_{3}$ solution has a normality of 0.05 eq/L.

#### Explanation:

This is a less than problem.

${\text{Ba"("NO"_3)_2 + "Na"_2"SO"_4 → "BaSO"_4 + "2NaNO}}_{3}$

Find the limiting reactant

(a) Calculate the initial moles of each reactant

"Initial moles of Ba"("NO"_3)_2 = 0.200 color(red)(cancel(color(black)("L"))) × ("0.35 mol Ba"("NO"_3)_2)/(1 color(red)(cancel(color(black)("L")))) = "0.0700 mol Ba"("NO"_3)_2

${\text{Initial moles of Na"_2"SO"_4 = 0.300 color(red)(cancel(color(black)("L"))) × ("0.2 mol Na"_2"SO"_4)/(1 color(red)(cancel(color(black)("L")))) = "0.060 mol Na"_2"SO}}_{4}$

(b) Calculate the moles of ${\text{NaNO}}_{3}$ from each reactant.

From "Ba"("NO"_3)_2:

${\text{Moles of NaNO"_3 = 0.0700 color(red)(cancel(color(black)("mol Ba"("NO"_3)_2))) × "2 mol NaNO"_3/(1 color(red)(cancel(color(black)("mol Ba"("NO"_3)_2)))) = "0.140 mol NaNO}}_{3}$

From ${\text{Na"_2"SO}}_{4}$:

${\text{Moles of NaNO"_3 = 0.060 color(red)(cancel(color(black)("mol Na"_2"SO"_4))) × "2 mol NaNO"_3/(1 color(red)(cancel(color(black)("mol Na"_2"SO"_4)))) = "0.12 mol NaNO}}_{3}$

${\text{Na"_2"SO}}_{4}$ is the limiting reactant, because it gives the fewest moles of ${\text{NaNO}}_{3}$.

Calculate the normality of the ${\text{NaNO}}_{3}$

At this point we have 0.12 mol ${\text{NaNO}}_{3}$ in 0.5 L of solution.

If we add another 2 L of water, we have 2.5 L of solution.

$\text{Molarity" = "moles"/"litres" = "0.12 mol"/"2.5 L" = "0.048 mol/L}$

Since ${\text{1 mol NaNO"_3 = "1 eq NaNO}}_{3}$,

$\text{Normality = 0.05 eq/L}$