Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chemistry 11 Chapter 6 STOICHIOMETRY OF EXCESS QUANTITIES.

Similar presentations


Presentation on theme: "Chemistry 11 Chapter 6 STOICHIOMETRY OF EXCESS QUANTITIES."— Presentation transcript:

1

2 Chemistry 11 Chapter 6 STOICHIOMETRY OF EXCESS QUANTITIES

3 Introduction: So far... b we have assumed that a given reactant is completely used up during the reaction

4 In reality... b reactions are often carried out in such a way that one or more of the second reactants actually are present in EXCESS amounts.

5 Definitions b EXCESS REACTANT = the reactant in excess b LIMITING REACTANT = the reactant that completely reacts b THE LIMITING REACTANT determines the yield of the product (how much product(s) will form)

6 A Simple Analogy b Imagine you work at McDonalds™ … b You have 10 hamburger buns and 5 beef patties b How many regular hamburgers can you make?

7 Da Answer... b Indeed, you would get 5 regular hamburgers! b And what was left over?

8 Connecting the Lingo … b There would be 5 hamburger buns in EXCESS! b Therefore, the beef patty is known as the LIMITING ingredient since it “limits” or determines how many regular buns can be made!

9 Note that b we do not predict based on the number of hamburger buns

10 Example 1 Example 1 b If 20.0 g of hydrogen gas react with g of oxygen, which reactant is present in excess and by how many grams?

11 Step 1 … b The balanced equation:

12 Step 2 … b First PREDICT which reactant is limiting (it’s ok if you predict wrong) b USUALLY the reactant with the least number of moles is limiting (but not always)

13 Convert masses to moles Number of moles of H 2 present = 20.0 g x 1 mol H 2 10 mol H 2 = 20.0 g x 1 mol H 2 = 10 mol H g H g H 2 Number of moles of O 2 present = g x 1 mol O 2 = g O 2 = g x 1 mol O 2 = mol O g O 2 Let’s make a prediction...

14 Prediction: O 2 is limiting Mass of H 2 that reacts with g O 2 = g O 2 x 1 mol O 2 x 2 mol H 2 x 2.0 g H g O 2 1 mol O 2 1 mol H 2 = g O 2 x 1 mol O 2 x 2 mol H 2 x 2.0 g H g O 2 1 mol O 2 1 mol H 2 = 12.5 g H 2 = 12.5 g H 2

15 Analysing the numbers … b What we have: g O 2 and 20.0 g H 2 b We predict O 2 b We predict O 2 is limiting (i.e. all g reacted) b b We calculated that we would need 12.5 g H 2 b Is the prediction correct ?

16 Da Answer (again!) b Yes!! Prediction is correct b only 12.5 g H 2 is required, so we have an excess of 7.5 g H 2 (20.0 g g) b so H 2 is in EXCESS of 7.5 g.

17 The other side of the coin So what if we predicted that H 2 was limiting? Mass of O 2 that reacts with 20.0 g H 2 = 20.0 g H 2 x 1 mol H 2 x 1 mol O 2 x 32.0 g O g H 2 2 mol H 2 1 mol O g H 2 2 mol H 2 1 mol O 2 = g O 2

18 Therefore … b If ALL 20.0 g of H 2 were to completely react we would need g of O 2 b b BUT we only have g of O 2 b b So the prediction that H 2 limiting is INCORRECT!

19 Example 2. If 79.1 g of Zn reacts with 1.05 L of 2.00 M HCl, a) Which reactant is in excess and by how much? b) What is the mass of each product?

20 a) which reactant is excess? The balanced equation: Zn + 2HCl  ZnCl 2 + H g 1.05 L, 2.00 M x g y g  mol 1.21 mol 2.10 mol (what we HAVE)

21 Prediction: Zn is limiting Moles of HCl required =1.21 mol Zn x =1.21 mol Zn x 2 mol HCl = 2.43 mol HCl 1 mol Zn Therefore 2.42 mol HCl would be required to react with 1.21 mol Zn. We ONLY have 2.10 mol HCl So is our prediction correct?

22 Uh Oh! You’re wrong! b We would need more HCl (2.42 mol) than what we have (2.10 mol) if all the Zn were to react b Thus: Zn is in excess, and HCl is limiting!

23 To find how much in excess: We must find how many moles of Zn is required to react with 2.10 molHCl Mol of Zn = 2.10 mol HCl x 1 molZn 2 molHCl = 1.05 mol Zn Excess Zn = = 0.16 mol Zn

24 b) mass of products? Since HCl is limiting we MUST use this amount to calculate the mass of products x g ZnCl 2 = 2.10 mol HCl x 1 mol ZnCl 2 x g 2 mol HCl 1 mol ZnCl 2 = 143 g ZnCl 2 y g H 2 = 2.10 mol HCl x 1 mol H 2 x 2.0 g 2 mol HCl 1 mol H 2 = 2.1 g H 2

25 Example 3: b 3.00 L of 0.1 M NaCl reacts with 2.50 L of M AgNO 3. Calculate the yield of solid AgCl (in grams) that will be produced. b This problem requires us to determine how much product (AgCl) will form, so we will need to first determine which reactant is limiting.

26 The balanced equation: NaCl (aq) + AgNO 3(aq) NaCl (aq) + AgNO 3(aq)  NaNO 3(aq) + AgCl (s) 3.00 L 2.50 L ? g 3.00 L 2.50 L ? g 0.1M M 0.1M M     mol mol Limiting Excess(since 1:1 ratio) Limiting Excess(since 1:1 ratio)

27 NaCl limiting... Therefore: mol NaCl = mol AgCl = mol (also 1:1 ratio) (also 1:1 ratio) Mass of AgCl = mol AgCl x g AgCl 1 mol AgCl = 43.1 g AgCl

28 Percent Yield b Often 100% of the expected amount of product cannot be obtained from a reaction b The term “Percent Yield” is used to describe the amount of product actually obtained as a percentage of the expected amount

29 Reasons for reduced yields A) the reactants may not all react because: i) not all of the pure material actually reacts actually reacts ii) the reactants may be impure impure B) Some of the products are lost during procedures such as solvent extraction, filtration etc

30 The equation: Percent Yield = ACTUAL YIELD x 100% THEORETICAL YIELD THEORETICAL YIELD b Actual yield = amount of product obtained (determined experimentally) b Theoretical yield = amount of product expected (determined from calculations based on the stoichiometry of the reaction) b The amounts may be expressed in g, mol, molecules

31 Types of calculations A) Find the percentage yield, given the mass of reactant used and mass of product formed B) Find the mass of product formed, given the mass of reactant used and the percentage yield C) Find the mass of reactant used, given the mass of product formed and percentage yield b Note that the percentage yield must be less than 100%  But when calculating the theoretical yield assume a 100% yield

32 Example 1 When 15.0 g of CH 4 is reacted with an excess of Cl 2 according to the reaction: CH 4 + Cl 2  CH 3 Cl + HCl a total of 29.7 g of CH 3 Cl is formed. Calculate the percentage yield.

33 The solution... The actual yield of CH 3 Cl = 29.7 g To find the theoretical yield of CH 3 Cl: (assuming a 100% yield) g of CH 3 Cl = 15.0 g CH 4 x 1 mol CH 4 g of CH 3 Cl = 15.0 g CH 4 x 1 mol CH 4 x 1 mol CH 3 Cl x 50.5 g 16.0g CH 4 1 mol CH 4 1 mol CH 3 Cl 16.0g CH 4 1 mol CH 4 1 mol CH 3 Cl = g = g

34 Then: Percentage yield = actual yield x 100% theoretical yield theoretical yield = 29.7 g x 100% = 62.7 % = 29.7 g x 100% = 62.7 % g g

35 Example deux! What mass of K 2 CO 3 is produced when 1.50 g of KO 2 is reacted with an excess of CO 2 if the reaction has a 76.0% yield? The reaction is: 4KO 2(s) + 2 CO 2(g)  2K 2 CO 3(s) + 3O 2(g)

36 The solution: We are looking for the actual yield (some idiot forgot to weigh and record the mass of product!) First calculate the mass of K 2 CO 3 produced (assuming a 100% yield) i.e. the theoretical yield g of K 2 CO 3 = 1.50 g KO 2 x 1 mol KO 2 x 2 mol K 2 CO 3 x g 71.1 g KO 2 4 mol KO 2 1 mol K 2 CO 3 = g actual yield = 76.0 % x g = x = 1.11 g K 2 CO 3

37 Last (but not least) example... What mass of CuO is required to make 10.0 g of Cu according to the reaction 2NH 3 + 3CuO  N 2 + 3Cu 2NH 3 + 3CuO  N 2 + 3Cu + 3H 2 O if the reaction has 58.0 % yield?

38 Here we go again … Actual yield = 10.0 g Cu From the percentage yield equation, calculate the theoretical yield of Cu. Theoretical yield of Cu = Actual yield x 100% Percentage yield = 10.0 g x 100 % 58.0 % = g

39 Now find the mass of CuO: Use this theoretical yield and find the mass of CuO that would be needed: g CuO = g Cu x 1 mol Cu x 3 mol CuO x 79.5g 63.5 g Cu 3 mol Cu 1 mol CuO = 21.6 g CuO


Download ppt "Chemistry 11 Chapter 6 STOICHIOMETRY OF EXCESS QUANTITIES."

Similar presentations


Ads by Google