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 thelimiting 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 thelimiting 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 limitingreagent.
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.