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#?
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
This tells you that the reaction will always consume equal numbers of moles of sodium hydroxide,
Now, theoretically, i.e. for a reaction that has a
However, in practice, chemical reactions never hit that
Your job here will to figure out the difference between the theoretical yield of the reaction, which is what you'd get at
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
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
However, it only produced
#"% yield" = (4.0 color(red)(cancel(color(black)("moles"))))/(5.0 color(red)(cancel(color(black)("moles")))) xx 100 = color(green)("80. %")#