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#?

1 Answer
Jan 10, 2016

Answer:

#"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. %")#