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Sometimes called limiting reactants.

Balanced Equations• A balanced chemical equation gives us theideal ratio of **reactants** and products.• In the real world, this ideal ratio is seldomachieved.• There is usually too much of one reactantand not enough of the other.• One of them is a **limiting** reagent.

A Simple Example• Hot dog buns come in packages of 8.• Hot dogs come in packages of 12.• If this is true, how long will it take agrasshopper with a wooden leg to kick allthe seeds out of a dill pickle?• Wait, that’s not right the right question.

A Simple Example• If you have one package of hot dogs (12dogs) and 2 packages of buns (16 buns),how many hot dogs in buns can you make?• Right, 12. The number of hot dogs is the**limiting** factor.• There are excess hot dog buns.

Limiting Reagents• A **limiting** reagent is **called** this because itlimits how much product can form.• Let’s use a familiar example:• 2H 2 + O 2 2 H 2 O

Limiting Reagents• What happens if there are more than 2hydrogen molecules? How much watercan we make now?

Limiting Reagents• Right! We can still only make 2 watermolecules. There will be hydrogenmolecules left over.

Limiting Reagents• The O 2 limits how much water can form. Itis the **limiting** reagent.• There is more H 2 than can react with theoxygen, so the hydrogen is in excess.• There are 2 water molecules with 4 H 2molecules left over.

Limiting Reagents• The concept works with moles, too.• If we started with 6 moles of H 2 and only 1mole of O 2 , we can only make 2 moles ofH 2 O.• There will be 4 moles of H 2 left over.• Oxygen is **limiting**, hydrogen is in excess.

Limiting Reagents• There is an Excel spreadsheet program witha link on my web page that will help toreinforce this concept.• http://staffweb.psdschools.org/rjensen/genchem/stoich.xls• Let’s play with it a bit.

Limiting Reagents• The way these problems are mostcommonly presented looks like this:• A balanced equation is given along with themass of 2 of the **reactants**.• For example: 2H 2 + O 2 2 H 2 O• If there are 10.0 g of H 2 and 10.0 g of O 2 ,which is the **limiting** reagent, and howmuch water will form?

An Example Problem• This is just 2 stoichiometry problems rolled intoone.• Convert the 10 g of each reactant to moles. (Step1 of the normal stoichiometry process.)• 10.0 g H 2 x (1 mol/2.0 g) = 5.0 moles H 2• 10.0 g O 2 x (1 mol/32.0 g) = 0.312 moles O 2• Select one of the materials and use the mole ratiofrom the equation to see if there is enough of theother material. (Step 2 of the process.)

An Example Problem• 5.0 mol H 2 x (1 mol O 2 /2 mol H 2 ) = 2.5 molO 2 . 2.5 mol O 2 are needed to react all theH 2 .• We already said there are only 0.312 molO 2 , which is not enough to react 5 moles ofH 2 .• Therefore, O 2 is the **limiting** reagent and H 2is in excess.

An Example Problem• The amount of product depends on the**limiting** reagent.• 0.312 mol O 2 will all be used up to makewater.• 0.312 mol O 2 (2 mol H 2 O/1 mol O 2 ) = 0.624mol H 2 O.• 0.624 mol H 2 O x (18.0 g/mol) = 11.2 g H 2 O.

Another Way to Solve• An alternative way to solve **limiting** reagentproblems is to solve using both startingamounts for the product in question.• The results will show 2 different answers.• The smaller of the 2 answers will be thecorrect one, and that reactant is the **limiting**reagent.

Another Way to Solve• 10.0g O 2 1 mol O 2 2 mol H 2 O 18.0g H 2 O = 11.2g H 2 O32.0 g O 2 1 mol O 2 1 mol H 2 O• 10.0g H 2 1 mol H 22 mol H 2 O 18.0g H 2 O = 90.0g H 2 O2.0 g H 2 2 mol H 2 1 mol H 2 O• Since we started with only 20 g of **reactants**,making 90 g of water is impossible! Thesmaller answer must be correct.• So O 2 must be **limiting**, and we get 11.2gH 2 O.