How do you identify the limiting reagent for the given combination of balanced reactions? #(NaOH)/(5.0 mol) + (HCl)/(4.5 mol) -> NaCl + H_2O#

1 Answer
Jul 5, 2017


Hydrochloric acid is the limiting reagent here.


The limiting reagent is simply the reactant that gets completely consumed before all the moles of the other reactant get the chance to take part in the reaction.

You know that sodium hydroxide and hydrochloric acid react in a #1:1# mole ratio.

#"NaOH"_ ((aq)) + "HCl"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O"_ ((l))#

This means that the reaction will always consume equal numbers of moles of each reactant, regardless of exactly how many moles of sodium hydroxide and of hydrochloric acid are available.

In your case, you know that #5.0# moles of sodium hydroxide react with #4.5# moles of hydrochloric acid. Since you don't have enough moles of hydrochloric acid to ensure that all the moles of sodium hydroxide react

#overbrace("5.0 moles HCl")^(color(blue)("what you need for 5.00 moles NaOH")) " " > " " overbrace("4.5 moles HCl")^(color(blue)("what you have available"))#

you can say that hydrochloric acid will act as a limiting reagent.

The reaction will consume #4.5# moles of hydrochloric acid--the limiting reagent is completely consumed--and #4.5# moles of sodium hydroxide.

Moreover, the reaction will produce #4.5# moles of aqueous sodium chloride and #4.5# moles of water, all on account of the #1:1# mole ratios that exist across the board.

After the reaction is finished, you will also be left with

#overbrace("5.0 moles NaOH")^(color(blue)("what you start with")) - overbrace("4.5 moles NaOH")^(color(blue)("what is consumed")) = "0.5 moles NaOH"#

This is why you can say that sodium hydroxide is in excess.