1. Limiting Reactions and Percent Yield Calculating by moles or mass ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

2. The reactant in a chemical reaction that is used up first It limits the amount of product Can find the limiting reactant based on MOLES or MASS ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

3. Finding the Limiting Reactant based on Mole 1. Reactants always combine according to the number of moles given in a balanced equation. 2. The limiting reactant is found by comparing the given moles to moles in the balanced equation. 3. Multiply the given moles by a mole ratio. ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

4. Example : 2H 2 + O 2 2H 2 O suppose you combine given: 4 mol of H 2 and 3 mol of O 2 1. 3 mol O 2 X 2 mol H 2 = 6 mol H 2 is needed 1 mol O 2 to react with 3 mol O 2 6 mol H 2 is needed and only given 4 mol H 2 2. 4 mol H 2 X 1 mol O 2 = 2 mol O 2 needed to 2 mol H 2 react with 4 mol H 2 ( limiting reactant will be H 2 used up in equation #2 and not enough in equation #1) ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

5. Example #2: if you combine 4 mol N 2 with 6 mol of H 2, what is the limiting reactant? Use this balanced equation: N 2 + 3H 2 2NH 3 4 mol N 2 X 3 mol H 2 = 12 mol H 2 needed to 1 mol N 2 react with 4 mol N 2 6 mol H 2 X 1 mol N 2 = 2 mol N 2 needed to 3 mol H 2 react with 6 mol H 2 The first equation is not possible you do not have 12 mol H 2. In second equation, all 6 mol H 2 was used. So, both equations show H 2 is the limiting reactant. It is all used up! ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

6. Finding the Limiting Reactant based on Mass Given in grams It is easier to determine the PRODUCT Mass that can be formed form each reactant mass Compare the 2 product masses and the reactant that produces the less product is the limiting reactant Use mass to mass (a reactant given to a product in grams) ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

7. EXAMPLE: You combine 75.0g of NH 3 with 120.0g of O 2. What is the limiting reactant? Balanced equation: 4 NH 3 + 7O 2 4NO 2 + 6H 2 O(pick one product) STEP #1: convert each mass into grams of a product. (I pick H 2 O!!) 75.0g NH 3 X 1 mole NH 3 X 6 mol H 2 O X 18.0g/mol H 2 O = 17.0g NH 3 4 mol NH 3 119g H 2 O ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

8. 120.0g O 2 X 1 mol O 2 X 6 mol H 2 O X 18.0g/mol H 2 O 32.0gO 2 7 mol O 2 = 57.9g H 2 O 57.9g H 2 O is lower than 119g H 2 O so the limiting reactant is O 2 !!!!! ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

9. Percent Yield Compares the amount of actual product to the amount predicted by stoichimetry Actual yield: the amount of product measured in lab Theoretical yield: the ideal amount of product predicted by stoichiometry % yield: actual yield X 100 theoretical yield ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

10. Example of % yield 87.3g of NaOH reacts according to the equation, H 2 SO 4 + 2NaOH Na 2 SO 4 + 2H 2 O There is an excess of H 2 SO 4. If 32.5g of water is actually produced, what is the % yield for water? 87.3gNaOHX1molNaOHX2molH 2 OX18.0gH 2 O 40.0gNaOH 2molNaOH 1molH 2 O = 39.3g H 2 O ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright

11. Percent yield of H 2 O % yield H 2 O = 32.5g X 100% 39.3g = 82.7% H 2 O ©2011 University of Illinois Board of Trustees http://islcs.ncsa.illinois.edu/copyright