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

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, SiO2, in the reaction. Would there be any of either reactant left over? SiO2 (s) + 2C (s) → Si (l) + 2CO (g)

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 UO2F2 by the reaction UF6 + 2H2O → UO2F2 + 4HF

Simple everyday example bicycle

2H2 (g) + O2 (g) 2H2O (g)

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

C + O2 → CO2

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

16000 g C 1mole C =1333 mol C 1mol Si = 667mol Si 12 g 2mol C How much silicon can be produced from 16 kg of carbon and 32 kg of silica, SiO2, in the reaction. Is there an excess of either reactant? SiO2(s) + 2C(s) → Si(l) + 2CO(g) 16000 g C 1mole C =1333 mol C 1mol Si = 667mol Si 12 g 2mol C 32000 g SiO2 1mole SiO2 =532 mol SiO2 1mol Si = 60.1 g 1mol SiO2 532 mol Si Excess Limiting reactant

532 moles SiO2 2 moles C = 1064 moles C used up 1 mole SiO2 Carbon is the excess reagent. Silicon dioxide is the limiting reagent? SiO2(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 SiO2 2 moles C = 1064 moles C used up 1 mole SiO2 Subtract moles of excess used from moles that you had. 1333 moles C – 1064 used = 269 moles left over. Convert to grams if necessary using molar mass.

3 C2H3O2H + Al(OH)3 → Al(C2H3O2)3 + 3 H2O If a student mixes 7 moles of C2H3O2H with 7 moles of Al(OH)3 how many moles of H2O can be made? 7 moles C2H3O2H 3 moles H2O = 7 moles H2O 3 moles C2H3O2H 7 moles Al(OH)3 3 moles H2O = 21 moles H2O 1 mole Al(OH)3 Problem 1 from Worksheet 10 Limiting reactant Excess

Subtract moles of excess used from moles you had. Al(OH)3 is the excess reagent. C2H3O2H is the limiting reagent? 3 C2H3O2H + Al(OH)3 → Al(C2H3O2)3 + 3 H2O Use the moles of limiting reagent and the mole ratio of reactants to determine how many moles of excess reactant are used up. 7moles C2H3O2H 1 moles Al(OH)3 =2.33molesAl(OH)3 3 moles C2H3O2H 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.

Problem 6 from Worksheet 10 If a student mixes 100 grams of K with 100 grams of Cl2 how much KCl could be made? 2K + Cl2→ 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 Cl21mole Cl2 = 1.4 mol Cl2 2mol KCl =2.8mol KCl 70.9 g 1mol Cl2 Problem 6 from Worksheet 10 Limiting reactant Excess

2.56 mol K 1 moles Cl2 = 1.28 moles Cl2 used up 2 mole K Chlorine is the excess reagent. Potassium is the limiting reagent? 2K + Cl2→ 2KCl Use the moles of limiting reagent and the mole ratio of reactants to determine how many moles of excess reactant are used up. 2.56 mol K 1 moles Cl2 = 1.28 moles Cl2 used up 2 mole K Subtract moles of excess used from moles that you had. 1.4 moles Cl2 – 1.28 used = 0.12 moles left over. Convert to grams if necessary using molar mass.