Chem107_Sect508_Lect7_final

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Instructor: Prof. Michael B. Hall Office Hrs: T 5:10 – 6:00 Outside ILCB 226 for General Questions Th 5:10 – 6:00 In ILCB 207 for General Questions M 5:00 – 6:00 By Zoom for Individual Conferences (1 st one next M) Chemistry Connection Point: Mon - Fri 8:00 am - 5:00 pm, ILSQ 3rd Floor Lobby REMINDERS & NEW ANNOUNCEMENTS: OWL homework (HW) due Sunday, 9/17, at 10pm. Good practice for Quiz. Next Quiz - Monday 9/18 on Unit 2 – Chemical Reactions (last week and this week) This quiz covers more material, so you should increase your study time. ARE YOU HAVING LAB THIS WEEK? HAVE YOU DONE THE LABQUIZ? DO YOU KNOW THE DRESS CODE? DO YOU HAVE GOOGLES? DO YOU HAVE A NOTEBOOK? Chemistry 107 - Section 508 Fall 2023 – Lecture #7
Reaction Stoichiometry Mass to Mass Conversions 1. Converting from grams of one chemical species to grams of another chemical species in a reaction.
2A + C ® 3B + D Reaction Stoichiometry Chp. 4.2
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Dehydration of Sucrose Demonstration When sucrose, C 12 H 22 O 11 , is dehydrated pure carbon forms. Write a balanced reaction for the dehydration of sucrose and calculate the mass of carbon formed if 80 g of sucrose is dehydrated. A. 2.80 g B. 6.66 g C. 33.68 g D. 0.23 g E. 2.57 g https://www.youtube.com/watch?v=GuHViMhYbRc&feature=youtu.be C 12 H 22 O 11 (s) ® 12 C (s) + 11 H 2 O (g)
Combustion in Automobiles Chp. 3, Chp. 4 incomplete combustion in air (80% N 2 , 20% O 2 ) C 8 H 18 (g) + N 2, O 2 (g) ® NO x (g) + CO(g) + C x H y O z (g) + CO 2 (g) + H 2 O(g) toxins and pollutants octane air ?
Stoichiometry: Gram to Gram Conversions Chp. 4.2 When 10.5 g of acetaldehyde (MW 44.05 g/mol) reacts with excess oxygen in a palladium catalyzed oxidation, how many grams of water (MW 18.01 g/mol) are formed? 2C 2 H 4 O (g) + 5O 2 (g) ® 4CO 2 (g) + 4H 2 O(g) A. 4.29 g B. 8.59 g C. 17.2 g D. 10.5 g
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Reaction Stoichiometry Limiting Reagents 1. Identifying limiting reagents in chemical reactions. 2. Converting from grams of one chemical species to grams of another chemical species in a reaction where there is a limiting reagent. 3. Determining how much of an excess reagent remains.
Limiting Reagent Concept Chp. 4.3
Limiting Reagent – reactant that is the first to be completely consumed in a reaction. Excess Reagent – any reactant that remains once the limiting reagent has been completely consumed. 2H 2 (g) + O 2 (g) ® 2H 2 O(l) Limiting Reagents Chp. 4.3
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Mg ( s ) + 2HCl ( aq ) → H 2 ( g ) + MgCl 2 ( aq ) 1.5 g of magnesium is reacted with 30 mL (Flask 1), 62.5 mL (Flask 2) and 90 mL (Flask 3) of 2.0 M HCl. Limiting Reagent Concept Demo https://youtu.be/wbgoNGAzoKo
Mg ( s ) + 2HCl ( aq ) → H 2 ( g ) + MgCl 2 ( aq ) 1.5 g of magnesium is reacted with 30 mL (Flask 1), 62.5 mL (Flask 2) and 90 mL (Flask 3) of 2.0 M HCl. 0.062 mol Mg 0.062 mol HCl 0.062 mol Mg 0.124 mol HCl 0.062 mol Mg 0.186 mol HCl 1 2 3 What is the limiting reagent in Reaction 3? A. Mg B. HCl C. Neither reactant is limiting Limiting Reagent Concept Demo
2A + C ® 3B + D Limiting Reagent Reaction Stoichiometry Chp. 4.3 Grams A Grams C Moles A Moles C Moles Limiting Reactant Moles B Grams B
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2NH 3 (g) + CO 2 (g) ® (NH 2 ) 2 CO(aq) +H 2 O(l) How many grams of urea, (NH 2 ) 2 CO, are formed when 637.2 g of NH 3 are reacted with 1142 g of CO 2 ? How much of the excess reactant remains? Limiting Reagents Chp. 4.3 1. Determine the limiting reagent by comparing moles NH 3 to moles CO 2 . 2. Use the moles (or grams to moles) of the limiting reagent to determine the moles then grams of urea formed. 3. Use the moles (or grams to moles) of the limiting reagent to determine the moles then grams of excess reagent reacted. 4. Subtract the grams of excess reagent reacted from the initial grams of excess reagent to determine the remaining excess reactant.
2NH 3 (g) + CO 2 (g) ® (NH 2 ) 2 CO(aq) +H 2 O(l) What is the limiting reagent when 637.2 g of NH 3 are allowed to react with 1142 g of CO 2 to form urea as a product? How many grams of urea, (NH 2 ) 2 CO, are formed? Limiting Reagents Chp. 4.3 Determine the limiting reagent by comparing moles NH 3 to moles CO 2 . Use the moles of the limiting reagent to determine the moles then grams of urea formed.
2NH 3 (g) + CO 2 (g) ® (NH 2 ) 2 CO(aq) +H 2 O(l) How many grams of the excess reagent remains? Limiting Reagents Chp. 4.3 Use the moles (or grams to moles) of the limiting reagent to determine the moles then grams of excess reagent reacted. Subtract the grams of excess reagent reacted from the initial grams of excess reagent to determine the remaining excess reactant.
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Stoichiometry: Limiting Reagents Chp. 4.3 Imagine that 20.0 g of octane (MW 114.23 g/mol) is reacted with 50.0 g oxygen (MW 31.98 g/mol), what is the limiting reagent? 2C 8 H 18 (g) + 25O 2 (g) ® 16CO 2 (g) + 18H 2 O(g) A. octane B. oxygen Is this a rich or lean fuel environment? A. lean B. rich
Stoichiometry: Limiting Reagents Chp. 4.3 Imagine that 20.0 g of octane (MW 114.23 g/mol) is reacted with 50.0 g oxygen (MW 31.98 g/mol), how many grams of carbon dioxide (MW 44.01 g/mol) are formed? 2C 8 H 18 (g) + 25O 2 (g) ® 16CO 2 (g) + 18H 2 O(g) A. 61.64 g B. 7.71 g C. 68.8 g D. 44.0 g
Stoichiometry: Limiting Reagents Chp. 4.3 Imagine that 20.0 g of octane (MW 114.23 g/mol) is reacted with 50.0 g oxygen (MW 31.98 g/mol), how much of the excess reagent is lef t over ? 2C 8 H 18 (g) + 25O 2 (g) ® 16CO 2 (g) + 18H 2 O(g) A. 5.74 g B. 35.8 g C. 30.0 g D. 14.3 g
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Reaction Stoichiometry Percent Yield 1. Defining and determining actual and thoeretical yields for chemical reactions. 2. Defining and calculating the percent yield for a chemical reaction. Percent Yield = x 100 Actual Yield Theoretical Yield
Percent Yield Chp. 4.4 What is the theoretical yield (maximum possible yield) when you pop 100 kernels of popcorn? What is the actual yield in the example above? What percentage of the kernels popped?
Percent Yield Chp. 4.4 What is the theoretical yield (maximum possible yield) when you pop 100 kernels of popcorn? A. 18 kernals B. 82 kernals C. 100 kernals
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