# What is the percent yield if 4.0 moles of NaCl are obtained when 5.0 moles of NaOH react with 6.0 moles of HCl in the reaction NaOH + HCl -> NaCl + H_2O?

Jan 10, 2016

$\text{80. %}$

#### Explanation:

Start by taking a look at the balanced chemical equation for this neutralization reaction

${\text{NaOH"_text((aq]) + "HCl"_text((aq]) -> "NaCl"_text((aq]) + "H"_2"O}}_{\textrm{\left(l\right]}}$

Notice that you have $1 : 1$ mole ratios between all the chemical species that take part in the reaction.

This tells you that the reaction will always consume equal numbers of moles of sodium hydroxide, $\text{NaOH}$, and hydrochloric acid, $\text{HCl}$.

Now, theoretically, i.e. for a reaction that has a 100% yield, one mole of hydrochloric acid will react with one mole of sodium hydroxide and produce one mole of aqueous sodium chloride, $\text{NaCl}$, and one mole of water.

However, in practice, chemical reactions never hit that 100% yield mark.

Your job here will to figure out the difference between the theoretical yield of the reaction, which is what you'd get at 100% yield, and the actual yield of the reaction.

In essence, a reaction's percent yield is a measure of how efficient that reaction is at converting reactants to products.

$\textcolor{b l u e}{\text{% yield" = "actual yield"/"theoretical yield} \times 100}$

Now, another important thing to notice here is that you don't have equal numbers of moles of sodium hydroxide and hydrochloric acid.

More specifically, you have

• $5.0$ moles of sodium hydroxide
• $6.0$ moles of hydrochloric acid

Since you have fewer moles of sodium hydroxide, this compound will be completely consumed by the reaction. The $1 : 1$ mole ratio tells you that $5.0$ moles of hydrochloric acid are needed to neutralize that many moles of sodium hydroxide.

Therefore, sodium hydroxide will act as a limiting reagent, i.e. it will determine how much acid will actually react.

So, you now know that the reaction consumed $5.0$ moles of $\text{NaOH}$ and $5.0$ moles of $\text{HCl}$. Theoretically, it should have produced $5.0$ moles of $\text{NaCl}$.

However, it only produced $4.0$ moles of $\text{NaCl}$. This means that its percent yield was

"% yield" = (4.0 color(red)(cancel(color(black)("moles"))))/(5.0 color(red)(cancel(color(black)("moles")))) xx 100 = color(green)("80. %")