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Chemistry Unit 8 - Stoichiometry

Chemistry Unit 8 - Stoichiometry. Lesson 2 – The Mole Ratio Book Section: 12.2 Objective: SWBAT solve mole-mole stoichiometry problems. Agenda: Lesson, Practice Problems, HW. The Key to Stoichiometry – The Mole Ratio. 2Al 2 O 3(l)  4Al (s) + 3O 2(g)

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Chemistry Unit 8 - Stoichiometry

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  1. ChemistryUnit 8 - Stoichiometry Lesson 2 – The Mole Ratio Book Section: 12.2 Objective: SWBAT solve mole-mole stoichiometry problems. Agenda: Lesson, Practice Problems, HW

  2. The Key to Stoichiometry – The Mole Ratio • 2Al2O3(l) 4Al(s) + 3O2(g) • Mole ratios allow us to “convert” between different reactants and products. • What is the mole ratio between Al2O3 and Al? • 2 mol Al2O3 : 4 mol Al • We can write this as a conversion factor: • We can now “convert” between amounts of different compounds.

  3. One Conversion: mol  mol • CO2(g) + 2LiOH(s) Li2CO3(s) + H2O(l) • How many moles of lithium hydroxide are required to react with 20 mol CO2? • Molar ratio between CO2 and LiOH? • 1:2 • = 40 mol LiOH

  4. You try: mol  mol • How many moles of chlorine gas (Cl2) would react with 5 moles of sodium (Na) according to the following chemical equation? (Balance equation) • Na + Cl2  NaCl • 2.5 mol Cl2

  5. HW 8-2 Due Wednesday • This week: • Wednesday: Mass-Mass Stoichoimetry (12.2) • Thursday: Limiting Reagents (12.3) • Friday: Percent Yield (12.3)

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