Question #a6b5e

Start by writing the balanced chemical equation that describes this neutralization reaction

#"H"_ 2"SO"_ (4(aq)) + color(blue)(2)"NaOH"_ ((aq)) -> "Na"_ 2"SO"_ (4(aq)) + 2"H"_ 2"O"_ ((l))#

Notice that the reaction consumes #color(blue)(2)# moles of sodium hydroxide for every mole of sulfuric acid that takes part in the reaction.

In other words, the reaction will always consume twice as many moles of sodium hydroxide than moles of sulfuric acid, regardless of how many moles of sulfuric acid take part in the reaction.

Now, you know that you have

• three moles of sulfuric acid, #3 xx "H"_2"SO"_4#
• four moles of sodium hydroxide, #4 xx "NaOH"#

All you have to do to figure out is you're dealing with a limiting reagent is pick one of the reactants and check to see if you have enough moles of the second reactant available.

For example, let's pick sulfuric acid. You know that #3# moles of sulfuric acid would require

#3 color(red)(cancel(color(black)("moles H"_2"SO"_4))) * (color(blue)(2)color(white)(a)"moles NaOH")/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) = "6 moles NaOH"#

Since you only have #4# moles of sodium hydroxide available, you can say that sodium hydroxide will act as a limiting reagent, i.e. it will be completely consumed before all the moles of sulfuric acid get the chance to react.

More specifically, the reaction will consume all the moles of sodium hydroxide and

#4color(red)(cancel(color(black)("moles NaOH"))) * ("1 mole H"_2"SO"_4)/(color(blue)(2)color(red)(cancel(color(black)("moles NaOH")))) = "2 moles H"_2"SO"_4#

After the reaction is complete, you will be left with #1# mole of unreacted sulfuric acid, since

#overbrace("3 moles H"_2"SO"_4)^(color(purple)("what you start with")) - overbrace("2 moles H"_2"SO"_4)^(color(darkgreen)("what is consumed")) = overbrace("1 mole H"_2"SO"_4)^(color(brown)("unreacted"))#