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Limiting and Excess Reactants  Lesson Essential Questions: How does the limiting reactant affect the percentage yield and actual yield in a chemical reaction?

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Presentation on theme: "Limiting and Excess Reactants  Lesson Essential Questions: How does the limiting reactant affect the percentage yield and actual yield in a chemical reaction?"— Presentation transcript:

1 Limiting and Excess Reactants  Lesson Essential Questions: How does the limiting reactant affect the percentage yield and actual yield in a chemical reaction? Vocabulary: limiting reactant, excess reactant, theoretical yield, actual yield, percent yield

2 Why? For example, a business manager, budgeting for the production of silicon ‑ based computer chips, wants to know how much silicon can be produced from 16 kg of carbon and 32 kg of silica, SiO 2, in the reaction. Would there be any of either reactant left over?  SiO 2 (s) + 2C (s) → Si (l) + 2CO (g)

3 Another example  A safety engineer in a uranium processing plant, wants to know how much water needs to be added to 25 pounds of uranium hexafluoride to maximize the synthesis of UO 2 F 2 by the reaction  UF 6 + 2H 2 O → UO 2 F 2 + 4HF

4 Simple everyday example bicycle

5 2H 2 (g) + O 2 (g) 2H 2 O (g)

6 Limiting Reactants b Limiting Reactant used up in a reaction determines the amount of product b Excess Reactant added to ensure that the other reactant is completely used up

7 C + O 2 → CO 2

8 Limiting Reactants 1. Write a balanced equation. 2. Convert the reactants to moles. 3. For each reactant, use the mole ratio of reactant to product and calculate the moles of product formed. 4. Smaller answer indicates: limiting reactant amount of product that will be formed

9 How much silicon can be produced from 16 kg of carbon and 32 kg of silica, SiO 2, in the reaction. Is there an excess of either reactant? SiO 2(s) + 2C (s) → Si (l) + 2CO (g) g C 1mole C =1333 mol C 1mol Si = 667mol Si 12 g 2mol C g SiO 2 1mole SiO 2 =532 mol SiO 2 1mol Si = 60.1 g 1mol SiO mol Si Limiting reactant Excess

10 Carbon is the excess reagent. Silicon dioxide is the limiting reagent? SiO 2(s) + 2C (s) → Si (l) + 2CO (g) Use the moles of limiting reagent and the mole ratio of reactants to determine how many moles of excess reactant are used up. 532 moles SiO 2 2 moles C = 1064 moles C used up 1 mole SiO 2 Subtract moles of excess used from moles that you had moles C – 1064 used = 269 moles left over. Convert to grams if necessary using molar mass.

11 3 C 2 H 3 O 2 H + Al(OH) 3 → Al(C 2 H 3 O 2 ) H 2 O  If a student mixes 7 moles of C 2 H 3 O 2 H with 7 moles of Al(OH) 3 how many moles of H 2 O can be made? 7 moles C 2 H 3 O 2 H 3 moles H 2 O = 7 moles H 2 O 3 moles C 2 H 3 O 2 H 7 moles Al(OH) 3 3 moles H 2 O = 21 moles H 2 O 1 mole Al(OH) 3 Problem 1 from Worksheet 10 Limiting reactant Excess

12 Al(OH) 3 is the excess reagent. C 2 H 3 O 2 H is the limiting reagent? 3 C 2 H 3 O 2 H + Al(OH) 3 → Al(C 2 H 3 O 2 ) H 2 O Use the moles of limiting reagent and the mole ratio of reactants to determine how many moles of excess reactant are used up. 7moles C 2 H 3 O 2 H 1 moles Al(OH) 3 =2.33molesAl(OH) 3 3 moles C 2 H 3 O 2 H used up Subtract moles of excess used from moles you had. 7 moles Al(OH) 3 – 2.33molesAl(OH) 3 = 4.67 moles left over. Convert to grams if necessary using molar mass.

13 If a student mixes 100 grams of K with 100 grams of Cl 2 how much KCl could be made? 2K + Cl 2 → 2KCl Convert both to moles. Then use mole ratio to predict the moles of product each could make. 100 g K 1mole K = 2.56 mol K 2mol KCl =2.56 mol KCl 39.1 g 2mol K 100g Cl 2 1mole Cl 2 = 1.4 mol Cl 2 2mol KCl =2.8mol KCl 70.9 g 1mol Cl 2 Problem 6 from Worksheet 10 Limiting reactant Excess

14 Chlorine is the excess reagent. Potassium is the limiting reagent? 2K + Cl 2 → 2KCl Use the moles of limiting reagent and the mole ratio of reactants to determine how many moles of excess reactant are used up mol K 1 moles Cl 2 = 1.28 moles Cl 2 used up 2 mole K Subtract moles of excess used from moles that you had. 1.4 moles Cl 2 – 1.28 used = 0.12 moles left over. Convert to grams if necessary using molar mass.


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