Limiting Reactions and Percent Yield Calculating by moles or mass ©2011 University of Illinois Board of Trustees

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

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

Example : 2H 2 + O 2 2H 2 O suppose you combine given: 4 mol of H 2 and 3 mol of O 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 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

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

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

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

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

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

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

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