Question

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`?

Answer 1

`"80. %"`

Explanation:

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

`"NaOH"_text((aq]) + "HCl"_text((aq]) -> "NaCl"_text((aq]) + "H"_2"O"_text((l])`

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, `"NaOH"`, and hydrochloric acid, `"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, `"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.

`color(blue)("% yield" = "actual yield"/"theoretical yield" xx 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 `"NaOH"` and `5.0` moles of `"HCl"`. Theoretically, it should have produced `5.0` moles of `"NaCl"`.

However, it only produced `4.0` moles of `"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. %")`