1. Sec 12.3 limiting reactant, percent, actual and theoretical Yield

2. Excess Reactant- the reactant that is leftover in a chemical reaction.
Definitions:
Limiting Reactant- the reactant that gets used up in a chemical reaction.
Excess Reactant- the reactant that is leftover in a chemical reaction.
Analogy:
When making brownies you have plenty of flour, but run out of eggs. The flour is the ___________ reactant, and the eggs are the _____________ reactant.

3. Excess Reactant- the reactant that is leftover in a chemical reaction.
Definitions:
Limiting Reactant- the reactant that gets used up in a chemical reaction.
Excess Reactant- the reactant that is leftover in a chemical reaction.
Analogy:
When making brownies you have plenty of flour, but run out of eggs. The flour is the excess reactant, and the eggs are the limiting reactant.

4. To calculate a Limiting Reactant:
1. Convert grams of each reactant to grams of one of the products.
2. Compare the grams of product for each. Whichever gives LESS product is the limiting reactant.

5. Ex. Which reactant is limiting if 25. 0g of N2 and 5
Ex. Which reactant is limiting if 25.0g of N2 and 5.0 g of H2 are mixed together? N2 + H2  NH3
Convert g mol  mol  g
for each reactant
Balanced equation:

6. Ex. Which reactant is limiting if 25. 0g of N2 and 5
Ex. Which reactant is limiting if 25.0g of N2 and 5.0 g of H2 are mixed together? N2 + H2  NH3
Convert g mol  mol  g
for each reactant
Balanced equation:
N2 + 3H2  2NH3

7. N2 + 3H2  2NH3 25.0 gN2- g N2 mol N2  mol NH3  g NH3
25.0 g N2 x 1 mol N2 x 2 mol NH3 x g NH3
28.02 g N2 1 mol N mol NH3
= g NH3

8. N2 + 3H2  2NH3 = 28 g NH3 5.0 g H2- g H2 mol H2  mol NH3  g NH3
5.0 g H2 x 1 mol H2 x 2 mol NH3 x g NH3
2.02 g H mol H mol NH3
= 28 g NH3

9. 2. Compare the two amounts of NH3
Starting with 25.0 g N2 = 30.4 g NH3
Starting with 5.0 g H2 = 28 g NH3
limiting: excess.

10. 2. Compare the two amounts of NH3
Starting with 25.0 g N2 = 30.4 g NH3
Starting with 5.0 g H2 = 28 g NH3
limiting: H2 excess: N2

11. Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation:
2Cu(s) + S(s)  Cu2S(s)
What is the limiting reagent when 80.0 g Cu
reacts with 25.0 g S?

12. Theoretical Yield The maximum amount of product that can be produced.
It is rarely obtained because of experimental errors

13. Actual (experimental) Yield
This is the amount of product that is actually obtained.

14. Percent Yield

15. Ex 1: What is the percent yield if the theoretical yield of product is 45g , but when doing the experiment you only got 30g of product?

16. Ex 1: What is the percent yield if the theoretical yield of product is 45g , but when doing the experiment you only got 30g of product?
Actual Yield= 30 g
Theoretical Yield = 45 g
Percent Yield = (30g / 45g) x 100 = 66.67%

17. Ex 2: You perform a synthesis reaction to make water from oxygen and hydrogen. You expect to produce 25,000 grams of water, but only get 18,750 g H2O what is your percent yield?

18. Ex 2: You perform a synthesis reaction to make water from oxygen and hydrogen. You expect to produce 25,000 grams of water, but only get 18,750 g H2O what is your percent yield?
Actual Yield= 18,750 g
Theoretical Yield = 25,000g
Percent Yield = ?
X = 18,750 g x = 75%
25,000 g

19. Al (s) + I2 (s)  Al I3 (s) 2Al (s) + 3I2 (s)  2AlI3 (s)
Ex 3. Al (s) + I2 (s)  Al I3 (s) If we react 2.40 g I2, but only produce g Al I3, what is the percent yield?
Balance the equation:
Find out the Theoretical Yield of Al I3
Find the Percent Yield
Balance the Equation
Al (s) + I2 (s)  Al I3 (s)
2Al (s) + 3I2 (s)  2AlI3 (s)

20. Find the Theoretical Yield (yes, three step conversion)
g I2  mol I2 mol AlI3 g AlI3
2.40 g I2 x 1 mol I2 x 2 mol AlI3 x 407.7g AlI3
253.8 g I2 3 mol I mol AlI3
= 2.57g AlI3

21. Find the Percent Yield Actual Yield= 2.05 g Theoretical Yield = 2.57 g
(2.05g / 2.57g) x 100% = %