# Question 24c1f

Apr 14, 2015

For your titration, sulfuric acid will be in excess, i.e. you added insufficient sodium hydroxide.

${H}_{2} S {O}_{4 \left(a q\right)} + \textcolor{red}{2} N a O {H}_{\left(a q\right)} \to N {a}_{2} S {O}_{4 \left(a q\right)} + 2 {H}_{2} O$

Notice that you have a $1 : \textcolor{red}{2}$ mole ratio between sulfuric acid and sodium hydroxide, which implies that, regardless of how many moles of sulfuric acid you have, you need twice as many moles of sodium hydroxide in order for the base to not act as a limiting reagent.

Comparing the number of moles of sulfuric acid and of sodium hydroxide you add will tell how which one of the two is the limiting reagent and which one is the excess reactant.

Use the two solutions' molarity to determine the number of moles of each species added

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{{H}_{2} S {O}_{4}} = \text{0.155 M" * 65.5 * 10^(-3)"L" = "0.0102 moles }$ ${H}_{2} S {O}_{4}$

${n}_{N a O H} = \text{0.416 M" * 20.0 * 10^(-3)"L" = "0.00832 moles }$ $N a O H$

As you can see, you don't have enough moles of sodium hydroxide present. The titration is not complete, which means that sulfuric acid is in excess.

The number of moles of sulfuric acid that will react is

0.00832cancel("moles NaOH") * ("1 mole "H_2SO_4)/(color(red)(2)cancel("moles NaOH")) = "0.00416 moles"

You have an excess of sulfuric acid of

n_(H_2SO_4"excess") = 0.0102 - 0.00416 = "0.00604 moles"#

Since you have a $1 : 1$ mole ratio between sulfuric acid and sodium sulfate, the number of moles of the latter produced by the reaction will be equal to the number of moles of the former that react

${n}_{N {a}_{2} S {O}_{4}} = {n}_{{H}_{2} S {O}_{4}} = \text{0.00416 moles}$

As a result, this reaction will produce

$0.00416 \cancel{\text{moles") * "142.04 g"/(1cancel("mol")) = color(green)("0.591 g}}$ $\text{ } N {a}_{2} S {O}_{4}$

If all the moles of sulfuric acid would have reacted, the produced mass would have been

$0.0102 \cancel{\text{moles") * "142.04 g"/(1cancel("moles")) = color(green)("1.45 g}}$ $\text{ } N {a}_{2} S {O}_{4}$

This is your theoretical yield - the mass that would have been produced if the sodium hydroxide wouldn't have acted as a limiting reagent.

SIDE NOTE The percent yield will be

$\text{% yield" = "actual yield"/"theoretical yield} \cdot 100$

$\text{% yield" = "0.591 g"/"1.45 g" * 100 = "40.8%}$