1. This work has been released into the public domain by its author, Benjah-bmm27. This applies worldwide. In some countries this may not be legally possible; if so: Benjah-bmm27 grants anyone the right to use this work for any purpose, without any conditions, unless such conditions are required by law.public domainBenjah-bmm27

2. Analogy Recipe for a ham sandwich: 2 slices of bread, 2 slices of ham, 1 slice of cheese. How many complete sandwiches can be made with 4 slices of bread, 5 slices of ham, and 4 slices of cheese? http://commons.wikimedia.org/wiki/File%3AToast-1.jpg By Rainer Z... (Own work) [GFDL (http://www.gnu.org/copyleft/fdl.html), CC-BY-SA-3.0 (http://creativecommons.org/licenses/by-sa/3.0/) or CC BY-SA 2.5-2.0-1.0 (http://creativecommons.org/licenses/by-sa/2.5-2.0-1.0)], via Wikimedia Commonshttp://commons.wikimedia.org/wiki/File%3ASchinken-gekocht.jpg By Rainer Zenz (Own work) [GFDL (http://www.gnu.org/copyleft/fdl.html) or CC-BY-SA-3.0 (http://creativecommons.org/licenses/by-sa/3.0/)], via Wikimedia Commons Answer: Limiting ingredient: 2 sandwiches Excess ingredients: Bread Ham and cheese = + + + + =

3. In a chemical reaction, when one reactant runs out, the reaction stops The reactant that runs out first is the Limiting Reactant. The reactants that are left over are the Excess Reactants. http://wps.prenhall.com/wps/media/objects/165/169519/blb9ch0307.html 2H 2 (g) + O 2 (g)  2H 2 O (g)

4. 10 mol H 2 x 1 mol O 2 = 5 mol O 2 needed 2 mol H 2 You have 7 mol O 2, you need 5 mol O 2, thus, 7 mol O 2 – 5 mol O 2 = 2 mol O 2 left over, so… O 2 is the excess reactant, you have 7 mol O 2, but you only need 5 mol O 2, therefore…. H 2 is the limiting reactant You can start with either reactant (it doesn’t matter). Let’s start with H 2 …

5. 2H 2 (g) + O 2 (g)  2H 2 O (g) 7 mol O 2 x 2 mol H 2 = 14 mol H 2 needed 1 mol O 2 Thus, you have 10 mol H 2, not the 14 mol H 2 you need, so… H 2 is the limiting reactant O 2 is the excess reactant by default Or you could start with O 2 and get the same result …

6. The Concept You must use the limiting reactant to determine the theoretical yield, NOT the excess reactant. Since we usually deal with grams in lab, not moles, we will need to add some additional steps to convert grams to moles…

7. The Mechanics Most problems start with grams, so: Step 1: Convert the grams of each reactant to moles. Step 2: Use the moles of either reactant to calculate the moles of the other reactant needed. Step 3: Compare what you have to what you need to determine the limiting reactant and excess reactant. Step 4: Complete the stoichiometry calculations using the limiting reactant. Optional: You can also detemine how much of the excess reactant is used up with a mole ratio and subtract from what you have to get the left over reactant.

8. It’s not as difficult as it sounds. Let’s try one together: The most important commercial process for converting N 2 and H 2 to ammonia, NH 3, is the Haber Process: N 2 (g) + 3H 2 (g)  2 NH 3(g) What mass of NH 3 can be formed from 12.0 g H 2 and 84.0 g N 2 ?

9. Ok, now you try: A strip of zinc metal weighing 2.00 g is placed in an aqueous solution containing 2.50 g of silver nitrate. (a)Which reactant is limiting? (b) How many grams of Ag will form? (c) How many grams of the excess reactant will be left at the end of the reaction?