Presentation on theme: "Limiting Reactants. You have a lit candle You put a jar over the candle What will happen? It will go out Why? It runs out of oxygen."— Presentation transcript:
You have a lit candle You put a jar over the candle What will happen? It will go out Why? It runs out of oxygen
Limiting Reactants A chemical reaction will stop when you run out of one of your products Limiting reactant – limits the extent of the reaction. Determines the amount of product that is formed. It runs out first Excess reactant – left over reactant
Limiting Reactant Problems TWO A limiting reactant problem can be easily identified because you have TWO givens You are going to do 2 stoichiometry problems You are going to do 2 stoichiometry problems
Steps 1. Write and balance the chemical equation 2. Start with given #1 and go to desired product 3. Start with given #2 and go to the same product 4. The one that formed the least amount of grams is the limiting reactant
Example S 8 + 4Cl 2 4S 2 Cl 2 S 8 + 4Cl 2 4S 2 Cl g of S g of Cl 2 S 2 Cl g of S 8 and g of Cl 2 are combined in a flask. How much S 2 Cl 2 will you get? Start with given # g S 2 Cl g S 8 x 1 mol S 8 x 4 mol S 2 Cl 2 x g S 2 Cl 2 = g S 2 Cl g S 8 1 mol S 8 1 mol S 2 Cl 2
Example Now do given # g S 2 Cl g Cl 2 x 1 mol Cl 2 x 4 mol S 2 Cl 2 x g S 2 Cl 2 = g S 2 Cl g Cl 2 4 mol Cl 2 1 mol S 2 Cl 2 Now we have a decision to make. Did we make g S 2 Cl 2 or g S 2 Cl g. The answer is g. After that amount, the reaction stopped S 8 + 4Cl 2 4S 2 Cl 2 S 8 + 4Cl 2 4S 2 Cl g of S 8 and g of Cl 2 are combined in a flask. How much S 2 Cl 2 will you get?
Other questions What was the limiting reactant? Cl 2 Cl 2 What was the excess reactant? S 8 S 8 How much excess did we have left over after the reaction was completed?
Other questions Before we can determine how much excess was left over, we must figure out how much was used. There are several ways to do this. Using stoichiometry, start with how much product you formed OR the total amount of the limiting reactant and go to the amount (g) of excess
Other questions This will tell you how much excess was used Now you subtract how much you used from how much you started with to give you the amount left over
Other questions In the previous example g S g Cl 2 x 1 mol Cl 2 x 1 mol S 8 x g S 8 = g S g Cl 2 4 mol Cl 2 1 mol S 8 S 8 This tells you how much S 8 was used. We want to know how much was left over g S 8. We started with g S 8 and we used g S g S 8 Therefore, we must have g S 8 left over
Another Example P 4 + 5O 2 P 4 O 10 P 4 + 5O 2 P 4 O 10 If we used 25.4 g P 4 and 50.0 g O 2 answer the following questions. 1.What is the limiting reactant? 2.What is the excess reactant? 3.How much P 4 O 10 will you get? 4.How much excess did you use? 5.How much excess will you have left over?
Another Example 1. What is the limiting reactant? P 4 2. What is the excess reactant?O 2 3. How much P 4 O 10 will you get? g P 4 O How much excess did you use? g O 2 used 5. How much excess will you have left over? g O 2 left over
% Yield Remember % = (part / whole) x 100 % = (part / whole) x 100 When performing an experiment, things do not always go exactly perfect Some product may get spilled, some may get sneezed on, or the reaction may not have gone to completion
% Yield Theoretical yield Theoretical yield –amount of product you should get if the experiment went perfectly. You get this number from stoichiometry Actual Yield Actual Yield – this is what you actually got in the lab. You measure this on a balance % yield % yield – how close you were to the correct answer % yield = (actual / theoretical) x 100 % yield = (actual / theoretical) x 100
% Yield Example K 2 CrO 4 + 2AgNO 3 Ag 2 CrO 4 + 2KNO 3 K 2 CrO 4 + 2AgNO 3 Ag 2 CrO 4 + 2KNO 3 What is the theoretical yield of Ag 2 CrO 4 formed from g AgNO g AgNO 3 x 1 mol AgNO 3 x 1 mol Ag 2 CrO 4 x g Ag 2 CrO 4 = g g AgNO 3 2 mol AgNO 3 1 mol Ag 2 CrO 4 Ag 2 CrO 4 What is the % yield if g is actually formed? What is the % yield if g is actually formed? % Yield = (actual / theoretical) x 100 % Yield = (actual / theoretical) x % % Yield = (0.455 / 0.488) x 100 = 93.2%