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Chemistry Week 26

Chemistry Week 26. Please get out your calculator!. 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.

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Chemistry Week 26

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  1. Chemistry Week 26 Please get out your calculator!

  2. 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 • Stoichiometry and Limiting Reactant. • If you missed Friday’s Quiz you must come make it up by Friday.

  3. Monday, March 11th Objective: Students will be able to determine a limiting reactant and calculate a percent yield. Bell Ringer: 2 C4H10 + 13 O2 8 CO2 + 10H2O How many grams of H2O are formed from 5.5 g of O2? 5.5g O2x ______ x ______ x ______ = g H2O

  4. 30 seconds Remaining…

  5. Monday, March 11th Objective: Students will be able to determine a limiting reactant and calculate a percent yield. Bell Ringer: 2 C4H10 + 13 O2 8 CO2 + 10H2O How many grams of H2O are formed from 5.5 g of O2? 5.5g O2x ______ x ______ x ______ = g H2O

  6. 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 • Stoichiometry and Limiting Reactant. • If you missed Friday’s Quiz you must come make it up by Friday.

  7. Monday, March 11th Objective: Students will be able to determine a limiting reactant and calculate a percent yield. Bell Ringer: 2 C4H10 + 13 O2 8 CO2 + 10 H2O How many grams of H2O are formed from 5.5 g of O2? 5.5g O2 x 1mol O2x 10molH2Ox 18 g H2O= 32 g O2 13 mol O2 1mol H2O 2.4g H2O =5.5 x 1 x 10 x 18 g H2O= 32 x 13 x 1

  8. 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. • How many bikes can be made if you have: • 18 wheels • 10 handlebars • 12 seats • 10 frames Example • You can make 9 bikes • Wheels are the limiting reactant.

  9. Example #1 • All limiting reactant problems start with a balanced chemical equation. 4 FeCl3 + 3 O2  2 Fe2O3 + 6 Cl2 How many moles of Cl2 can be produced if 5 moles of FeCl3 react with 4 moles of O2? • FeCl3 is the limiting reactant because 7.5 moles of Cl2 were produced. • O2 was the excess reactant. 5 mol FeCl3 x 6 mols Cl2 = 4 mols FeCl3 7.5 mol Cl2 4mol O2 x 6 mols Cl2= 3 moles O2 8mol Cl2

  10. Example #2 2 C2H6 + 7 O2  4 CO2 + 6 H2O How many grams of H2O can be produced if 15 grams of C2H6 react with 45 grams of O2? Molar masses: C2H6 = 30 g/mol; O2 = 32 g/mol; H2O = 18g/mol 18 gH2O = 1mol H2O 1mol C2H6 x 30 g C2H6 6 mol H2O x 2 mol C2H6 27 g H2O 15 g C2H6 x 21.7 g H2O 18 g H2O = 1mol H2O 6 mols H2Ox 4 moles O2 45 g O2 x 1 mol O2x 32 g O2 • O2 is the limiting reactant • C2H6 is the excess reactant

  11. Obj:SWBAT determine a limiting reactant and calculate a percent yield. Date: 3/11/2013 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) Percent Yield = Actual Value x 100 Theoretical Value Example What would be the percent yield of the previous reaction if only 20 g of H2O were produced? 92.2 % Percent Yield = 20 g H2O x 100 = 21.7 g H2O

  12. Set-Up Help Balanced ChemEq: aA bB • Calculate the molar masses of both compounds in the problem • Convert: remember Given and Want Given(g) x1 molGivenx molWant x Mol.massWant = Mol.massGivenmolGiven 1 molWant • Multiply everything in numerator. • Multiply everything in the denominator • Divide Numerator by Denominator.

  13. Set-Up Help Balanced ChemEq: aA bB • Mole A Mole B  Gram B: Mole A xb Mole B x Molar Mass B = Gram B a Mole A 1 molB • Gram A  Mole A  Mole B Gram A x 1Mole Ax b Mole B= Mole B Molar Mass A amolA

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