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Limiting Reagent, Excess Reagent, Theoretical Yield and % Yield DR. CHIN CHU.

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Presentation on theme: "Limiting Reagent, Excess Reagent, Theoretical Yield and % Yield DR. CHIN CHU."— Presentation transcript:

1 Limiting Reagent, Excess Reagent, Theoretical Yield and % Yield DR. CHIN CHU

2 H 2(g) H 2 O (g) O 2(g) + Not Balanced! +heat 2H 2(g) 2H 2 O (g) O 2(g) + +heat Balanced!

3 2H 2(g) 2H 2 O (g) O 2(g) + +heat Balanced Chemical Equation: 2 mole (H 2 )= 1 mole (O 2 )=2 mole (H 2 O) 4 g (H 2 ) = 32 g (O 2 ) = 36 g (H 2 O) Stoichiometric matched!

4 2H 2(g) 2H 2 O (g) O 2(g) + +heat Balanced Chemical Equation: 2.0 g (H 2 )12.0 g (O 2 )? g (H 2 O) Not stoichiometrically matched. Which reagent will run out first (limiting)? Which will have leftover (excess)?

5 2H 2(g) 2H 2 O (g) O 2(g) + +heat Step #1: Pick any reagent, calculate how much is needed for the stoichiometric amount of the other reagent. 2.0 g (H 2 )g (O 2 ) g (H 2 ) mole (H 2 ) mole (O 2 ) = 16.0 g (O 2 )

6 Step #2: Compare the stoichiometric amount of the other reagent with what is available. 2H 2(g) 2H 2 O (g) O 2(g) + +heat 2.0 g (H 2 )16.0 g (O 2 ) need to complete the reaction g (O 2 ) But only hasavailable. Hence, O 2 will run out first (limiting reagent), and H 2 will have leftover (excess reagent).

7 2H 2(g) 2H 2 O (g) O 2(g) + +heat Step #1: Pick a reagent, calculate how much is needed for the stoichiometric amount of the other reagent g (O 2 )g (H 2 ) g (O 2 ) mole (O 2 ) mole (H 2 ) = 1.5 g (H 2 ) Alternatively, pick the other reagent (O 2 ):

8 Step #2: Compare the stoichiometric amount of the other reagent with what is available. 2H 2(g) 2H 2 O (g) O 2(g) + +heat 1.5 g (H 2 )12.0 g (O 2 ) need to complete the reaction. 2.0 g (H 2 ) The system hasavailable. Hence, again O 2 will run out first (limiting reagent), and H 2 will have leftover (excess reagent). More than enough.

9 Question: How much of excess reagent is left after the reaction has stopped? 1.5 g (H 2 )is need to completely react with12.0 g (O 2 ). 2.0 g (H 2 ) is available. The amount in excess is: 2.0 g (H 2 ) What is available 1.5 g (H 2 ) Stoichiometric amount = 0.5 g (H 2 ) The amount in excess

10 Question: What is the maximum amount of water in grams that can be produced in this particular reaction (theoretical yield)? Which reagent to chose for the stoichiometry calculation? Logically, limiting reagent should be chosen for calculating theoretical yield since it will run out first in the reaction.

11 2H 2(g) 2H 2 O (g) O 2(g) + +heat Balanced Chemical Equation: 2.0 g (H 2 )12.0 g (O 2 )O 2 is the limiting reagent g (O 2 )g (H 2 O) g (O 2 ) mole (O 2 ) mole (H 2 O) = 13.5 g (H 2 O) Theoretical Yield

12 In most chemical reactions, the extent of how complete the reaction proceeds is less than 100%, leading to product yield less than the maximum amount possible (theoretical yield). % Yield= Actual Yield (g) Theoretical Yield (g) x 100%

13 Question: For the previously mentioned chemical reaction of hydrogen gas and oxygen gas, if the actual production of water is 10.8 grams, what is the % yield of the reaction? % Yield= Actual Yield (g) Theoretical Yield (g) x 100% 10.8 g 13.5 g = 80.0 %

14 Summary of the Process: 1.Balance the chemical equation. 2.Choose one reagent, calculate the stoichiometric amount the other reagent needed to complete reaction. 3.Compare the calculated stoichiometric amount with what is available to determine limiting reagent, excess reagent. Calculate the amount of leftover accordingly. 4.Use the limiting reagent, calculate theoretical yield for each product. 5.Calculate % yield based on actual yield and theoretical yield.


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