Agenda: 1/13/2017 Go over the procedure for the Molarity Lab Work on Day 1 of Molarity Lab Investigate the concept of Molarity Check Homework and Pre-Lab from last night Limiting Reactants and Percent Yield Summarize the concept of limiting reactants and develop an understanding of percent yield Practice questions on limiting reactants and percent yield Homework Quia: Limiting Reactants

Limiting Reactants and Percent Yield

Limiting Reactants

Vocabulary Limiting Reactant In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms. In other words– the stuff you run out of first Excess Reactant The reactant that is not completely used up in a reaction.

Steps Write and balance the reaction equation Convert given information to moles Use the mole ratio (from the balanced reaction) to determine which reactant will run out first. Once you determine the limiting reagent you can use that information to determine the maximum amount of product that can be made.

Example 79.1 g of zinc react with 63.5 g of HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Write the balanced reaction. Zn + 2 HCl -----------> H2 + ZnCl2 2. Covert the given information to moles 79.1 g Zn 1 mol Zn = 1.21 mol Zn 65.409 g Zn 63.5 g HCl 1 mol HCl = 1.74 mol HCl 36.46094 g HCl 3. Use the mole ratio to determine which reactant will run out first. 1.21 mol Zn 2 mol HCl = 2.42 mol HCl 1 mol Zn For 1.21 mol of Zn you would need 2.42 mol of HCl, but you only have 1.74 mol HCl, that means the HCl will run out first and be the limiting reagent

Example Continued How many liters of hydrogen are formed at STP? Since HCl is the limiting reagent we use the amount of moles that it has available to find the amount of H2 that is formed. 1.74 mol HCl 1 mol H2 = 0.870 mol H2 2 mol HCl Now we need to use Mole Island to convert from moles of H2 gas to liters of H2 gas. 0.870 mol H2 22.4 L = 19.5 L of H2 gas at STP 1 mol

Percent Yield

Equation Percent Yield = actual yield X 100 theoretical yield

Example When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. The actual amount produced in 46.3 g of KCl. K2CO3 + 2HCl  2KCl + H2O + CO2 Solve for the theoretical yield of KCl by setting up a stoichiometric conversion. 45.8 g of K2CO3 1 mole K2CO3 2 mol KCl 74.5513 g KCl 138.2055 g K2CO3 1 mol K2CO3 1 mol KCl = 49.4 g KCl

Example Continued 2. Solve for percent yield. Actual yield: 46.3g KCl Theoretical yield: 49.4g KCl Percent Yield = 46.3/49.4 X 100 = 93.7 %