1. Actual and Percent Yields So far when we have been talking about reactions we have been talking about 100% of the (limiting) reactant becoming a product. However in real- life most of the time some of the reactants do not react so our theoretical (if 100% of reactants become product) yield is different from our actual yield. We can express this difference as a percentage to see if a reaction would be viable.

2. LiOH + KCl  KOH + LiCl If I start this reaction with 20g of LiOH, what mass of LiCl should I get? If I actually get 6g of LiCl, what is my percent yield? Step 1: Write a balanced equation Step 2: Calculate the number of moles of reactant 1 mole LiOH = 24g 20g = 20/24 moles = 0.83 moles

3. Step 3: Use mole ratios to calculate the mass of product 1 mole LiOH : 1 mole LiCl 0.83 moles LiOH : 0.83 moles LiCl 1 mole LiCl = 42.5g 0.83 moles = 42.5 x 0.83 = 35.28g Theoretical Yield Step 4: Calculate percent yield using the formula % yield = Actual yield Theoretical yield X 1006 35.28 X 100 = 17% yield From question From step 3