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Announcements WebStudies test and Tutorial Tut Test 5 (formative assessment): Friday, 12 th 12:00 → Tuesday, 16 th 18:00 Tutorial 6: Thursday 18 th 14:00.

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Presentation on theme: "Announcements WebStudies test and Tutorial Tut Test 5 (formative assessment): Friday, 12 th 12:00 → Tuesday, 16 th 18:00 Tutorial 6: Thursday 18 th 14:00."— Presentation transcript:

1 announcements WebStudies test and Tutorial Tut Test 5 (formative assessment): Friday, 12 th 12:00 → Tuesday, 16 th 18:00 Tutorial 6: Thursday 18 th 14:00

2 the mole defining the mole molar mass amount-mass-number conversions mass percentage determine formula of unknown compound empirical/molecular formulas writing & balancing chemical equations calculating quantities (stoichiometry) mole-mole conversions from equations limiting reactants reaction yields fundamentals: solution stoichiometry molarity amount-mass-number conversions diluting molar solutions reactions in solution reading assignment: chapter 2 videos WebStudies

3 learning opportunities peer-assisted group learning (PAGL) interactive –style teaching enquiry based learning prepare for contact session: you do assigned reading you watch video clips Blackboard you work related sample problems text book/web you reflect: wonder, think about the concepts during contact session: we ask & test understanding of concepts we put up questions and discuss the content your group discusses/decides a response your scribe (guy in the middle) communicates your group compares group behind/front no-one calls answers, chews, misbehaves, etc.

4 the mole defining the mole molar mass amount-mass-number conversions mass percent determine formula of unknown compound empirical/molecular formulas writing & balancing chemical equations calculating quantities (stoichiometry) mole-mole conversions from equations limiting reactants reaction yields fundamentals: solution stoichiometry molarity amount-mass-number conversions diluting molar solutions reactions in solution this lecture

5 You make sandwiches that contain exactly 2 slices of bread and 1 slice of ham. If you have 23 slices of ham and 50 slices of bread, how many sandwiches can you make? limiting reactants a) b) c) d) 50 25 23 none of the above 2 slices bread + 1 ham → 1 sandwich 50 slices23 slices theory 2 bread 1 ham given 50 bread 23 ham given 2.17 bread 1 ham (theory) (given amounts)

6 limiting reactants a) b) c) d) You want to make sandwiches that specifically contain 2 slices of bread, 1 slice of ham and 3 slices of pickles. How many sandwiches can you make if you have 31 slices of bread, 17 slices of ham and 42 slices of pickles? 14 15 17 none of the above 2 bread + 1 ham + 3 pickles → 1 sandwich 311742 31 bread × 1 sandwich 2 bread 17 ham × 1 sandwich 1 ham 42 pickles × 1 sandwich 3 pickles = 15.5 sandwich = 17 = 14 limiting theory given amounts

7 a) b) c) d) limiting reactants Consider the reaction stated below. If you reacted 4 moles of N 2 and 6 moles of H 2, which is the limiting reactant? a)N 2 b)H 2 c)NH 3 d)none of the above N 2 (g) + 3 H 2 (g) 2 NH 3 (g) N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) stoichiometry: 3 mol H 2 1 mol N 2 given: 6 mol H 2 4 mol N 2 = 1.5 mol H 2 1 mol N 2 (theory) b) (theory) (given) limiting

8 A reaction vessel contains 6 moles of N 2 and 4 moles of H 2 that reacts according to the reaction stated below. Which is the limiting reactant? N 2 + 3 H 2 2 NH 3 N 2 + 3 H 2 → 2 NH 3 limiting reactants a) b) c) N2N2 H2H2 NH 3 NH 3 (mol): = 12 mol NH 3 NH 3 (mol): = 2.7 mol NH 3 3 H 2 1 N 2 theory 4 H 2 6 N 2 given 0.6 H 2 1 N 2 given limiting OR

9 a) b) c) d) e) limiting reactants Consider the reaction stated below. If you had 4 moles of N 2 and 6 moles of H 2, what is the maximum amount of NH 3 one could produce in this reaction? a)8 moles NH 3 b)4 moles NH 3 c)12 moles NH 3 d)6 moles NH 3 N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) NH 3 (mol): b) = 4 mol NH 3 = 8 mol NH 3

10 Asirvatham, 2010 Examine the molecular view of the reaction between AB and B 2 in the gas phase: The correct statement about this reaction is: a) The balanced equation: AB + B 2 → AB 3 b) AB and B 2 are initially in stoichiometric amounts c)AB is the limiting reactant d)The product of the reaction is A 2 B a) b) c) d) e) balanced equation:2 AB + B 2 → 2 AB 2 c) limiting reactants

11 excess limiting CH 4 + 2 O 2 → CO 2 + 2 H 2 O given: 2 mole O 2 5 mol CH 4 = want stoichiometry: 2 mole O 2 1 mol CH 4 have = 2121 0.4 1 Wood, 2006 Identify whether this presentation represents the combustion of methane (CH 4 ) with a) stoichiometric amount of oxygen b)methane limiting c) oxygen limiting c) a) b) c) b) c)

12 want have limiting excess CH 4 + 2 O 2 → CO 2 + 2 H 2 O given: 7 mole O 2 2 mol CH 4 stoichiometry: 2 mole O 2 1 mol CH 4 = 2121 = 3.5 1 Wood, 2006 Identify whether this presentation represents the combustion of methane (CH 4 ) with a) stoichiometric amount of oxygen b)methane limiting c) oxygen limiting c) a) b) c) b)

13 want have CH 4 + 2 O 2 → CO 2 + 2 H 2 O given: 6 mole O 2 3 mol CH 4 = stoichiometry: 2 mole O 2 1 mol CH 4 = 2121 Wood, 2006 Identify whether this presentation represents the combustion of methane (CH 4 ) with a) stoichiometric amount of oxygen b)methane limiting c) oxygen limiting 2121 c) a) b) c) b) a)

14 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) → 2 SO 3 (g) Given 3.0 moles of SO 2. O 2 would be the limiting reactant if we had less than ___ mol O 2 limiting reactants Given 5 moles of O 2. SO 2 would be the limiting reactant if we had less than ___ mol SO 2 Given 3.5 moles of O 2. SO 2 is in excess if we had more than ___ mol SO 2 Given 1.5 moles of SO 2. O 2 is in excess if we had more than ____ mol O 2 1.5 10 7.0 0.75 0.75 1.5 10 7.0 a) b) c) d) e) 3.5

15 If you have 8 moles of hydrogen gas and all the oxygen gas you need, how many moles of water can you make? a) b) c) d) e) a)1 mol b)2 mol c)4 mol d)8 mol e)16 mol calculations with equations O 2 + 2 H 2 → 2 H 2 O If you have all the hydrogen gas that you need and 8 moles of oxygen gas, how many moles of water can you make? If you have 8 moles of hydrogen gas and 9 moles of oxygen gas, how many moles of water can you make? stoich: 2 mol H 2 1 mol O 2 = 2121 given: 8 mol H 2 9 mol O 2 = 0.89 1 limiting

16 limiting reactants tip: whenever the mass or amount of both reactants are given, you can bet your bottom dollar that one of the reactants will be limiting challenging problem: sodium hydroxide reacts with carbon dioxide to give solid sodium carbonate and liquid water. when 1.70 mol sodium hydroxide is reacted with 1.00 mol carbon dioxide, the actual yield was found to be 93.3%. how many grams of sodium carbonate was produced?


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