Chemistry Week 26 Please get out your calculator!

March 12, 2013 AGENDA: 1 – Agenda/ Bell Ringer 2 – Pre-Lab Instructions 3 – Lab 4 – Clean-up and Post-Lab Today’s Goal: Students will be able to determine a limiting reactant and calculate a percent yield for a laboratory experiment. Homework 1. Limiting Reactant Lab – Calculations due tomorrow 2. Stoichiometry and Limiting Reactant 3. If you missed Friday’s Quiz you must come make it up by Friday.

Tuesday, March 12 th Objective: Students will be able to determine a limiting reactant and calculate a percent yield for a laboratory experiment. Bell Ringer: Predict the products of the following chemical equation and balance (see Pre-lab): ___HCl + ___Zn  ______ + ______ (*hint: Zn is a +2 charge)

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Tuesday, March 12 th Objective: Students will be able to determine a limiting reactant and calculate a percent yield for a laboratory experiment. Bell Ringer: Predict the products of the following chemical equation and balance (see Pre-lab): ___HCl + ___Zn  ______ + ______ (*hint: Zn is a +2 charge)

March 12, 2013 AGENDA: 1 – Agenda/ Bell Ringer 2 – Pre-Lab Instructions 3 – Lab 4 – Clean-up and Post-Lab Today’s Goal: Students will be able to determine a limiting reactant and calculate a percent yield for a laboratory experiment. Homework 1. Limiting Reactant Lab – Calculations due tomorrow 2. Stoichiometry and Limiting Reactant 3. If you missed Friday’s Quiz you must come make it up by Friday.

Tuesday, March 12 th Objective: Students will be able to determine a limiting reactant and calculate a percent yield for a laboratory experiment. Bell Ringer: Predict the products of the following chemical equation and balance (see Pre-lab): ___HCl + ___Zn  ___ ZnCl 2 + ___ H 2 (*hint: Zn is a +2 charge)

Lab Notes  We completed the Pre-Lab as the Bell Ringer. Please make sure you complete it while you are doing your lab.  You need to work quickly! Let each reaction occur for at least 7 minutes, or until all bubbles stop forming.  Make sure once you are done with a reaction, you quickly set up your next reaction. You want to leave as much time as possible for the reaction.  While you are waiting for the reaction to proceed, you can be setting up for your next trial by measuring out the HCl and Zn and doing the calculations.

1 st Period Lab Groups Amy Robert Arielle Mariama Tatiana Quay Hydeia Shana Patricia Sara Isaiah Niamarion Carla Gerald Tumani Chronicle Ashanti Choice Zahra Jane Imani Chibuzo Rahamanh Jasmine Danielle Kayla Kaylan Judy Maria Taylor Justin Joy

2 nd Period Lab Groups Chris Aginah Jenna Jared Regan Omari Ahzia Keira Keanthoy Joe Ajai Nathaly Camille Alan Javonte Jasmine Daija Sergio Ryan Tesia Otis Maya Najae Erick Angelique Kayla Lauryn Janiece

5 th Period Lab Groups Tyrese Courtney Romalle Tiara NeTanyah Tajana Darnell Destini Hunter Clarissa Kailah Jelani Dreshion Michael K Judah Jamille Kahlia Erik Savon Khamya Victoria Tanisha Jocelyn Khyri Badieye Michael G Murrell Chantice Breanna

6 th Period Lab Groups Jamie Tenisha Alex Julian Arya Cyntrall Nicole Jessica Jordin Michael Khalin D’angelo Erica Kyla Bria Teon Nanyamka Sydney Caylen D’niko Kihinde Ariel Tyler

8 th Period Lab Groups Keon Charles Abdul Patience Angela Ebony Brianna S Eric Paul Kennesha Nakia Sam Tayva Tamarha Carl Eriun Tiana Ravion Breanna R Jalaun Maya Sydney Willie Alexis Courtney Brittany David Elijah

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 O 2  8 CO 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 O 2  2 Fe 2 O 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 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 O 2  4 CO 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 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 7moles 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