## Presentation on theme: "Chemistry Week 26 Please get out your calculator!."— Presentation transcript:

March 11, 2013 AGENDA: 1 – Agenda/ Bell Ringer 2 – CN: Limiting Reagent and Percent Yield 3 – Summary 4 – Classwork Time Today’s Goal: Students will be able to determine a limiting reactant and calculate a percent yield. Homework 1. Stoichiometry and Limiting Reactant. 2. If you missed Friday’s Quiz you must come make it up by Friday.

Monday, March 11 th Objective: Students will be able to determine a limiting reactant and calculate a percent yield. Bell Ringer: 2 C 4 H 10 + 13 O 2  8 CO 2 + 10 H 2 O How many grams of H 2 O are formed from 5.5 g of O 2 ? 5.5g O 2 x ______ x ______ x ______ = g H 2 O

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Monday, March 11 th Objective: Students will be able to determine a limiting reactant and calculate a percent yield. Bell Ringer: 2 C 4 H 10 + 13 O 2  8 CO 2 + 10 H 2 O How many grams of H 2 O are formed from 5.5 g of O 2 ? 5.5g O 2 x ______ x ______ x ______ = g H 2 O

March 11, 2013 AGENDA: 1 – Agenda/ Bell Ringer 2 – CN: Limiting Reagent and Percent Yield 3 – Summary 4 – Classwork Time Today’s Goal: Students will be able to determine a limiting reactant and calculate a percent yield. Homework 1. Stoichiometry and Limiting Reactant. 2. If you missed Friday’s Quiz you must come make it up by Friday.

Monday, March 11 th Objective: Students will be able to determine a limiting reactant and calculate a percent yield. Bell Ringer: 2 C 4 H 10 + 13 O 2  8 CO 2 + 10 H 2 O How many grams of H 2 O are formed from 5.5 g of O 2 ? 5.5g O 2 x 1mol O 2 x 10molH 2 O x 18 g H 2 O = 32 g O 2 13 mol O 2 1mol H 2 O 2.4 g H 2 O = 5.5 x 1 x 10 x 18 g H 2 O = 32 x 13 x 1

Obj: SWBAT determine a limiting reactant and calculate a percent yield. Date: 3/11/2013 What is a Limiting Reactant? In a chemical reaction, the amount of product that can be produced is limited by the reactant that can produce the least amount of products based upon the ratios in which they react. The reactant that produces the least amount of product is called the limiting reactant. Example How many bikes can be made if you have: 18 wheels 10 handlebars 12 seats 10 frames You can make 9 bikes Wheels are the limiting reactant.

Example #1  All limiting reactant problems start with a balanced chemical equation. 4 FeCl 3 + 3 O 2  2 Fe 2 O 3 + 6 Cl 2 How many moles of Cl 2 can be produced if 5 moles of FeCl 3 react with 4 moles of O 2 ? 5 mol FeCl 3 x 6 mols Cl 2 = 4 mols FeCl 3 7.5 mol Cl 2 4 mol O 2 x 6 mols Cl 2 = 3 moles O 2 8 mol Cl 2 FeCl 3 is the limiting reactant because 7.5 moles of Cl 2 were produced. O 2 was the excess reactant.

Example #2 2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 O How many grams of H 2 O can be produced if 15 grams of C 2 H 6 react with 45 grams of O 2 ? Molar masses: C 2 H 6 = 30 g/mol; O 2 = 32 g/mol; H 2 O = 18g/mol 15 g C 2 H 6 x 1mol C 2 H 6 x 30 g C 2 H 6 6 mol H 2 O x 2 mol C 2 H 6 45 g O 2 x 1 mol O 2 x 32 g O 2 21.7 g H 2 O O 2 is the limiting reactant C 2 H 6 is the excess reactant 18 gH 2 O = 1 mol H 2 O 27 g H 2 O 6 mols H 2 O x 4 moles O 2 18 g H 2 O = 1 mol H 2 O

What is Percent Yield? The ratio of the actual yield (what you actually produced) to the theoretical yield (what you calculated) for a chemical reaction expressed as a percentage. It’s a measure of the efficiency of the reaction. (think of it like a grade for the reaction) Example Percent Yield = Actual Value x 100 Theoretical Value What would be the percent yield of the previous reaction if only 20 g of H 2 O were produced? Percent Yield = 20 g H 2 O x 100 = 21.7 g H 2 O 92.2 % Obj: SWBAT determine a limiting reactant and calculate a percent yield. Date: 3/11/2013

Set-Up Help Balanced Chem Eq: aA  bB 1. Calculate the molar masses of both compounds in the problem 2. Convert: remember Given and Want Given(g) x1 mol Given x mol Want x Mol.mass Want = Mol.mass Given mol Given 1 mol Want 3. Multiply everything in numerator. 4. Multiply everything in the denominator 5. Divide Numerator by Denominator.

Set-Up Help Balanced Chem Eq: aA  bB  Mole A  Mole B  Gram B: Mole A x b Mole B x Molar Mass B = Gram B a Mole A 1 mol B  Gram A  Mole A  Mole B Gram A x 1Mole A x b Mole B = Mole B Molar Mass A a mol A