1. 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

2. Limiting Reactants and Percent Yield

3. Limiting Reactants

4. 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.

5. 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.

6. 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 HCl > H ZnCl2
2. Covert the given information to moles
79.1 g Zn mol Zn = mol Zn
g Zn
63.5 g HCl mol HCl = mol HCl
g HCl
3. Use the mole ratio to determine which reactant will run out first.
1.21 mol Zn mol HCl = 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

7. 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 mol HCl 1 mol H2 = mol H2 2 mol HCl Now we need to use Mole Island to convert from moles of H2 gas to liters of H2 gas mol H L = 19.5 L of H2 gas at STP 1 mol

8. Percent Yield

9. Equation
Percent Yield = actual yield X 100 theoretical yield

10. 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 K2CO mole K2CO mol KCl g KCl
g K2CO mol K2CO mol KCl
= 49.4 g KCl

11. 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 %