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III. Stoichiometry Stoy – kee – ahm –eh - tree Chapter 12.

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1 III. Stoichiometry Stoy – kee – ahm –eh - tree Chapter 12

2 Sections Click the section to jump to the slides Mole Ratios Mole-to-Mole Calculations Mole-to-Mass Calculations Particle Calculations Molar Volume Calculations Limiting Reactants Percent Yields

3 Things you should remember From the Moles Unit: –Identify particles as atoms, molecules (mc), and formula units (fun) –1 mole = 6.02 x atoms, molecules, or formula units –1 mole substance = mass (in grams) from the periodic table From the Naming & Formulas Unit: –How to write a formula given a chemical name From the Chemical Reactions Unit: –How to write a chemical equation given words –Balancing equations

4 I. Stoichiometry ostoikheion, meaning element ometron, meaning measure oThus Stoichiometry- measuring elements!

5 III. Stoichiometry oStoichiometry (sometimes called reaction stoichiometry to specify its use in analyzing a chemical reaction) is the calculation of quantitative (numbers) relationships between the reactants and products in a balanced chemical reaction.

6 A. Basis for Calculations The basis for properly working stoichiometry problems is the Balanced Chemical Equation and the Mole Ratio. These are VITAL TO YOUR SURVIVAL IN STOICHIOMETRY!!!

7 1. Review of a Balanced Chemical Equation You must always check to ensure you have a proper chemical equation. It is not correctly balanced you can not use it for any calculations!!!!

8 Ex: Properly Balance the following equation __H 2 SO 4 (aq) + __NaHCO 3 (s)  __Na 2 SO 4 (aq) + __H 2 O(l) + __CO 2 (g) 2 2 2

9 2. Molar Ratio Once the equation is properly balanced, there are relationships between all of the compounds involved. These are called MOLE RATIOS. any two compounds can be written as a relationship in terms of moles

10 What’s the relationship between NaHCO 3 and Na 2 SO 4 ? 2 mole NaHCO 3 = 1 mol Na 2 SO 4 (it’s just the coefficients!!) What’s the relationship between H 2 SO 4 and CO 2 ? 1 mol H 2 SO 4 = 2 mol CO 2 (it’s just the coefficients!!) For the Reaction: H 2 SO 4 + 2NaHCO 3  Na 2 SO 4 + 2H 2 O + 2CO 2

11 1 mol H 2 SO 4 = 2 mol NaHCO 3 or 1 mc H 2 SO 4 = 2 mc NaHCO 3 or 1 mol/mc H 2 SO 4 2 mol/mc NaHCO 3 or 2 mol/mc NaHCO 3 1 mole/mc H 2 SO 4 For the Reaction: H 2 SO 4 + 2NaHCO 3  Na 2 SO 4 + 2H 2 O + 2CO 2

12 1 mol H 2 SO 4 = 2 mol CO 2 or 1 mc H 2 SO 4 = 2 mc CO 2 or 1 mol/mc H 2 SO 4 2 mol/mc CO 2 or 2 mol/mc CO 2 1 mole/mc H 2 SO 4 For the Reaction: H 2 SO 4 + 2NaHCO 3  Na 2 SO 4 + 2H 2 O + 2CO 2

13 Example: Iron reacts with oxygen to create iron(III) oxide. After writing a balanced equation, write down 3 possible relationships. (hint: just look at the coefficients.) Skeletal equation: Balanced equation: Fe + O 2  Fe 2 O 3 4Fe + 3O 2  2Fe 2 O 3 Relationships: 4 mol Fe = 3 mol O 2 4 mol Fe = 2 mol Fe 2 O 3 3 mol O 2 = 2 mol Fe 2 O 3

14 Practice 1. Write three mole ratios (relationships) from the reaction below: Al 2 S 3 + H 2 O  Al(OH) 3 + H 2 S 1 mol Al 2 S 3 = 6 mol H 2 O mol Al 2 S 3 = 2 mol Al(OH) 3 1 mol Al 2 S 3 = 3 mol H 2 S 6 mol H 2 O = 2 mol Al(OH) 3 6 mol H 2 O = 3 mol H 2 S 2 mol Al(OH) 3 = 3 mol H 2 S Y o u s h o u l d h a v e 3

15 Practice 1. Write three mole ratios (relationships) from the reaction below: Al 2 S 3 + H 2 O  Al(OH) 3 + H 2 S 1 mol Al 2 S 3 = 6 mol H 2 O mol Al 2 S 3 = 2 mol Al(OH) 3 1 mol Al 2 S 3 = 3 mol H 2 S 6 mol H 2 O = 2 mol Al(OH) 3 6 mol H 2 O = 3 mol H 2 S

16 Practice 2. Aluminum is produced by decomposing aluminum oxide into aluminum and oxygen. a. Write a balanced equation. b. Write all the molar ratios that can be derived from this equation. 2Al 2 O 3  4Al + 3O 2 2 mol Al 2 O 3 = 4 mol Al 2 mol Al 2 O 3 = 3 mol O 2 4 mol Al = 3 mol O 2

17 NOTE: NOTE: Every time you do a stoichiometric calculation you MUST use a mole ratio. The mole ratio allows you to compare one compound in an equation with another. Don’t forget to balance your chemical equations!!!

18 B. Calculating Problems

19 B. Calculating Problems - Stoich it up! Before any stoich problem you have to set it up. Consider this the pre-game warm- up. This should become second nature to you.

20 Pre-game Warm-up: 1. Write a balanced reaction. 2. Determine your given & want 3. Determine your relationships. If you see… 2 different substances, determine their mole ratio mass (g, mg, kg), calculate the molar mass of that substance atoms, molecules, or fun, 1 mol = 6.02 x of that type extras like mg or kg, you know what to do Now you are ready to solve- IT’S GAME TIME.

21 Game Time: 1.put your GIVEN OVER 1 2.place your relationships where the units cancel out diagonally 3.everything equal to each other goes above and below each other 4.cancel out your units until you are left over with your wanted

22 Here is a flow chart that we will dissect this unit to do our problems

23 For most of the examples we will be using this equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) Haber Process: an industrial process for producing ammonia from nitrogen and hydrogen by combining them under high pressure in the present of an iron catalyst source: worldnet.princeton.edu Haber Process:

24 1. Moles to Moles

25 Ex 1: How many moles of ammonia are produced from 4.00 moles of hydrogen gas in the presence of excess nitrogen? Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) mol H 2 3 mol H Given : Want : Relationships: 4.00 mol H 2 ? mol NH 3 2 mol NH 3 = 3 mol H 2 = mol NH 3 2 mol NH 3 x No grams, No molar mass!! No fun, mc, No 6.02 x !!

26 = mol N 2 Ex 2: How many moles of nitrogen were used if 7.8 moles of ammonia were made in excess hydrogen gas? Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) 1 mol N mol NH 3 2 mol NH Given : Want : Relationships: 7.8 mol NH 3 ? mol N 2 2 mol NH 3 = 1 mol N 2 X No grams, No molar mass!! No fun, mc, No 6.02 x !!

27 = moles H 2 Ex 3: How many moles of hydrogen react with 13 moles of nitrogen? Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) 1 13 mol N 2 1 mol N 2 39 Given : Want : Relationships: 13 mol N 2 ? mol H 2 1 mol N 2 = 3 mol H 2 3 mol H 2 x No grams, No molar mass!! No fun, mc, No 6.02 x !!

28 2. Moles to Mass/ Mass to Moles

29 Steps to Success! 1. Write a balanced reaction. 2. Determine your given & want 3. Determine your relationships. If you see… –2 different substances, determine their mole ratio –mass (g, mg, kg), calculate the molar mass of that substance –atoms, molecules, or f.un, 1 mol = 6.02 x of that type –extras like mg or kg, you know what to do

30 Ex 1: How many grams of ammonia are produced from 4.00 moles of hydrogen gas in excess nitrogen? Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) Given : Want : Relationships: 4.00 mol H 2 ? g NH 3 2 mol NH 3 = 3 mol H 2 How do I know when to use mole ratios, molar mass or 6.02 x ? Look at the “given” and “want” for clues 1 mol NH 3 = g NH 3 No fun, mc, No 6.02 x !!

31 = g NH 3 Ex 1: How many grams of ammonia are produced from 4 moles of hydrogen gas in excess nitrogen? mol H 2 3 mol H mol NH 3 x Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) Given : Want : Relationships: 4.00 mol H 2 ? g NH 3 2 mol NH 3 = 3 mol H 2 1 mol NH 3 = g NH 3 1 mol NH g NH 3 x

32 Ex 2: How many moles of ammonia are produced from 4.00 grams of hydrogen gas in excess nitrogen? Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) Given : Want : Relationships: 4.00 g H 2 ? mol NH 3 2 mol NH 3 = 3 mol H 2 Remember your Steps to Success! 1 mol H 2 = 2.016g H 2 No fun, mc, No 6.02 x !!

33 = mol NH 3 Ex 2: How many moles of ammonia are produced from 4 grams of hydrogen gas in excess nitrogen? g H g H mol H 2 x Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) 3 mol H 2 2 mol NH 3 x Given : Want : Relationships: 4.00 g H 2 ? mol NH 3 2 mol NH 3 = 3 mol H 2 1 mol H 2 = 2.016g H 2

34 Practice 1. How many grams of nitrogen will react with 3.40 moles of hydrogen to produce ammonia? 2. How many grams of hydrogen are required to make 54.0 moles of ammonia in excess nitrogen? 3. How many moles of nitrogen react completely with 3.70 moles of hydrogen? 31.7 g N g H mol N 2

35 3. Mass to Mass

36 Ex 1: How many grams of ammonia are produced from 4.00 grams of hydrogen gas in excess Nitrogen? Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) Given : Want : Relationships: 4.00 g H 2 ? g NH 3 2 mol NH 3 = 3 mol H 2 Remember your Steps to Success! 1 mol H 2 = 2.016g H 2 1 mol NH 3 = g NH 3

37 = g NH 3 Ex 1: How many grams of ammonia are produced from 4 grams of hydrogen gas in excess Nitrogen? g H g H mol H 2 x Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) 3 mol H 2 2 mol NH 3 x Given : Want : Relationships: 4.00 g H 2 ? g NH 3 2 mol NH 3 = 3 mol H 2 1 mol H 2 = 2.016g H 2 1 mol NH 3 = g NH 3 1 mol NH g NH 3 x

38 Ex 2: How many grams of Nitrogen react with 12.0 grams of Hydrogen? Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) Given : Want : Relationships: 12.0 g H 2 ? g N 2 Remember your Steps to Success! 1 mol H 2 = 2.016g H 2 1 mol N 2 = g N 2

39 = g N 2 Ex 2: How many grams of Nitrogen react with 12.0 grams of Hydrogen? g H g H mol H 2 x Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) 3 mol H 2 1 mol N 2 x Given : Want : Relationships: 12.0 g H 2 ? g N 2 1 mol N 2 = 3 mol H 2 1 mol H 2 = 2.016g H 2 1 mol N 2 = g N 2 1 mol N g N 2 x

40 Practice 1.Determine the mass of NH 3 produced from 280 g of N 2. 2.What mass of nitrogen is needed to produce 100. kg of ammonia? 340 g NH x 10 4 g N 2

41 3. MC/F.UN/ Atoms to Mass

42 Ex 1: How many grams of ammonia are produced from 2.09 x mc of hydrogen gas? Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) Given : Want : Relationships: 2.09 x mc H 2 ? g NH 3 2 mol NH 3 = 3 mol H 2 1 mol NH 3 = g NH 3 1 mol H 2 = 6.02 x mc H 2

43 Ex 1: How many grams of ammonia are produced from 2.09 x mc of hydrogen gas? Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) Given : Want : Relationships: 2.09 x mc H 2 ? g NH 3 2 mol NH 3 = 3 mol H 2 1 mol NH 3 = g NH 3 1 mol H 2 = 6.02 x mc H 2 = g NH x mc H x mc H x mol H 2 x 3 mol H 2 2 mol NH 3 x 1 mol NH g NH 3 x

44 Ex 2: How many molecules of ammonia are produced from 40.2 g of H 2 in the presence of excess N 2 ? Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) Given : Want : Relationships: 40.2g H 2 ? mc NH 3 2 mol NH 3 = 3 mol H 2 1 mol H 2 = 2.016g H 2 1 mol NH 3 = 6.02 x mc NH 3

45 Ex 2: How many molecules of ammonia are produced from 40.2 g of H 2 in the presence of excess N 2 ? = mc NH g H g H x mol H 2 x 3 mol H 2 2 mol NH 3 x 1 mol NH x mc NH 3 x Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) Given : Want : Relationships: 40.2g H 2 ? mc NH 3 2 mol NH 3 = 3 mol H 2 1 mol H 2 = 2.016g H 2 1 mol NH 3 = 6.02 x mc NH 3

46 Practice 1.How many mc of nitrogen will react with exactly 7.04 grams of hydrogen? 2.How many mc of ammonia will 2.09 x mc of N 2 produce in excess hydrogen? 7.01 x mc N x mc NH 3

47 1) How many mc of nitrogen will react with exactly 7.04 grams of hydrogen? = mc N g H g H x mol H 2 x 3 mol H 2 1 mol N 2 x 6.02 x mc N 2 x Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) Given : Want : Relationships: 7.04 g H 2 ? mc N 2 1 mol N 2 = 3 mol H 2 1 mol H 2 = 2.016g H 2 1 mol N 2 = 6.02 x mc N 2

48 How many mc of ammonia will 2.09 x mc of N 2 produce in excess hydrogen? Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) Given : Want : Relationships: 2.09 x mc N 2 ? mc NH 3 2 mol NH 3 = 1 mol N 2 1 mol NH 3 = 6.02 x NH 3 1 mol N 2 = 6.02 x mc N 2 = mc NH x mc N x mc N x mol N 2 x 2 mol NH 3 x 1 mol NH x mc NH 3 x

49 4. Molar Volume of Gases (Liters)

50 4. Molar Volume of Gases (liters) There is one new conversion: 1 mol of an ideal gas = 22.4 L of an ideal gas. For now we will assume every gas is an ideal gas at STP. You follow the same steps we have been following all along to solve this problem. In a reaction involving gases, Avogadro found that their volumes combine in whole number ratios. In other words, the coefficients in the chemical reaction also represent volume.

51 For gases: When you see liters… 1 mol __ = 22.4 L __

52 Ex 1: How many moles of nitrogen gas are needed to react with 44.8 liters of hydrogen gas to produce ammonia gas? Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) Given : Want : Relationships: 44.8 L H 2 ? mol N 2 1 mol N 2 = 3 mol H 2 Remember your Steps to Success! 1 mol H 2 = 22.4 L H 2 No grams, No molar mass No fun, mc, No 6.02 x !!

53 = mol N 2 Ex 1: How many moles of nitrogen gas are needed to react with 44.8 liters of hydrogen gas to produce ammonia gas? L H L H mol H 2 x Balanced Equation: N 2 (g) + 3H 2 (g)  2NH 3 (g) 3 mol H 2 1 mol N 2 x Given : Want : Relationships: 44.8 L H 2 ? mol N 2 1 mol N 2 = 3 mol H 2 1 mol H 2 = 22.4 L H 2 Given over 1 1 mol = 22.4 L Mole ratio

54 Ex 2: If 5.00 moles of H 2 react with excess N 2, how many liters of NH 3 are produced? N 2 (g) + 3H 2 (g)  2NH 3 (g) Given : Want : Relationships: 5.00 mol H 2 ? L NH 3 2 mol NH 3 = 3 mol H 2 Remember your Steps to Success! 1 mol NH 3 = 22.4 L NH 3 No grams, No molar mass No fun, mc, No 6.02 x !!

55 = L NH mol H 2 3 mol H mol NH 3 x 1 mol NH L NH 3 x Given over 1 Mole ratio 1 mol = 22.4 L Ex 2: If 5.00 moles of H 2 react with excess N 2, how many liters of NH 3 are produced? N 2 (g) + 3H 2 (g)  2NH 3 (g) Given : Want : Relationships: 5.00 mol H 2 ? L NH 3 2 mol NH 3 = 3 mol H 2 1 mol NH 3 = 22.4 L NH 3

56 II. Limiting Reactants and Excess Reactants

57 In reality, a scientist does not always add chemicals in perfect proportions. There is usually one reactant that is in “excess”. The limiting reactant or limiting reagent is the reactant that limits the amounts of the other reactants that can combine and the amount of product that can form in a chemical reaction. “which reactant are you going to run out of first?”

58 The substance that is not used up completely in a reaction is sometimes called the excess reactant (excess) Once one of the reactants is used up, no more product can be formed “which reactant do you have extra of?”

59 Which is the limiting reactant 4 shells 13 tires 1 shell 4 tires 1 car = 4 cars can be made if 4 shells are available = 3.25 cars can be made if 13 tires are available Less product can be produced, therefore tires must be the limiting reactant

60 Activity Write a recipe for the perfect burger

61 Ex. Consider the following balanced equation: Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g) If 1.21 moles of zinc are added to 2.64 moles of HCl, which reactant is in excess, and which is the limiting reactant? 1.21 mol Zn 2.64 mol HCl Available 2 mol HCl 1 mol Zn = 2.42 mol HCl = 1.32 mol Zn Needed

62 Ex. Consider the following balanced equation: Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g) If 1.21 moles of zinc are added to 2.64 moles of HCl, which reactant is in excess, and which is the limiting reactant? Available VS Needed 1.21 mol Zn 1.32 mol Zn 2.64 mol HCl 2.42 mol HCl < > Limiting reactant Excess reactant

63 To find the limiting reactant and the excess reactant: 1) Determine the balanced chemical equation 2) Convert the given/available amounts reactants to the number of moles of the other reactant(s) 3) Compare the available amounts to the needed amounts: *If the available amount > needed amount, it is excess reactant * If the available amount < needed amount, it is the limiting reactant 4) To calculate the amount of excess: -Available Amount – Needed Amount = Amount of Excess 5) To calculate the amount of product produced: -G: amount of available limiting reactant -W: amount of product

64 Example Some rocket engines use a mixture of hydrazine, N 2 H 4, and hydrogen peroxide, as the propellant. The products formed are nitrogen gas and steam. Which is the limiting reactant when 0.75 mol of N 2 H 4 is mixed with 0.50 mol of H 2 O 2 ? How much excess reactant, in moles, remains unchanged? How much of each product, in moles, is formed?

65 Some rocket engines use a mixture of hydrazine, N 2 H 4, and hydrogen peroxide, as the propellant. The products formed are nitrogen gas and steam. Step 1: Write a balanced reaction N 2 H 4 + H 2 O 2  N 2 (g)+ H 2 O(g) 4 2

66  Which is the limiting reactant when mol of N 2 H 4 is mixed with mol of H 2 O 2 ? N2H4N2H4 + H 2 O 2  N 2 (g)+ H 2 O(g) mol N 2 H mol H 2 O 2 Available 2 mol H 2 O 2 1 mol N 2 H 4 = 1.50 mol H 2 O 2 = 0.25 mol N 2 H 4 Needed E x c e s s L i m i t i n g R e a c t a n t

67 To find the limiting reactant and the excess reactant: 4. To calculate the amount of excess: -Available Amount – Needed Amount = Amount of Excess

68  How much excess reactant, in moles, remains unchanged? 0.75 mol N 2 H mol H 2 O 2 Available 2 mol H 2 O 2 2 mol H 2 O 2 = 1 mol N 2 H 4 1 mol N 2 H 4 = 1.50 mol H 2 O mol N 2 H 4 Needed Excess *Available – Needed = Remaining excess - = 0.50 mol N 2 H 4

69 To find the limiting reactant and the excess reactant: 5. To calculate the amount of product produced: -G: amount of available limiting reactant -W: amount of product

70 How many moles of product are formed? 0.50 mol H 2 O 2 G: W: R: 2 mol H 2 O 2 2 mol H 2 O 2 = 1 mol N 2 mole N 2 1 mol N 2 = 0.25 mol N 2 N2H4N2H4 + H 2 O 2  N 2 (g)+ H 2 O(g) 4 2 G: amount of available limiting reactant N 2 is the first product 0.50 mol H 2 O 2 G: W: R: 2 mol H 2 O 2 2 mol H 2 O 2 = 4 mol H 2 O mole H 2 O 4 mol H 2 O = 1.0 mol H 2 O G: amount of available limiting reactant H 2 O is the second product

71 Ex. Zinc metal and solid sulfur (S 8 ) react to form solid zinc sulfide. If 2.0 moles of zinc are heated with 2.0 mol S 8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed? 2.0 mol Zn 1.0 mol S 8 Available 1 mol S 8 8 mol Zn = 0.25 mol S 8 = 8.0 mol Zn Needed Balanced Equation: 8 Zn + S 8  8 ZnS

72 Ex. Zinc metal and solid sulfur (S 8 ) react to form solid zinc sulfide. If 2.0 moles of zinc are heated with 1.0 mol S 8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed? Available VS Needed 2.0 mol Zn 8.0 mol Zn 1.0 mol S mol S 8 < > Limiting reactant Excess reactant

73 How much excess is present? (Available Amount – Needed Amount = Amount of Excess) 1.0 mol S 8 – 0.25 mol S 8 = 0.75 mol S 8 in excess

74 To find the limiting reactant and the excess reactant: 4. To calculate the amount of excess: -Available Amount – Needed Amount = Amount of Excess 5. To calculate the amount of product produced: -G: amount of available limiting reactant -W: amount of product

75 Ex. Zinc metal and solid sulfur (S 8 ) react to form solid zinc sulfide. If two moles of zinc are heated with 1.00 mol S 8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed? 2 mol Zn G: W: R: 8 mol Zn 8 mol Zn = 8 mol ZnS mol ZnS 8 mol ZnS = 2 mol ZnS formed How much product was formed? Balanced Equation: 8 Zn + S 8  8 ZnS

76 Practice 1. Carbon reacts with steam at high temperatures to produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, identify the limiting reactant. How many moles of each product are formed? 2.4 mol C 3.1 mol H 2 O Available 1 mol H 2 O 1 mol C = 2.4 mol H 2 O = 3.1 mol C Needed Balanced Equation: C + H 2 O  H 2 + CO

77 Practice 1. Carbon reacts with steam at high temperatures to produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, identify the limiting reactant. How many moles of each product are formed? Available VS Needed 2.4 mol C 3.1 mol C 3.1 mol H 2 O2.4 mol H 2 O < > Limiting reactant Excess reactant

78 To find the limiting reactant and the excess reactant: 4. To calculate the amount of excess: -Available Amount – Needed Amount = Amount of Excess 5. To calculate the amount of product produced: -G: amount of available limiting reactant -W: amount of product

79 Carbon reacts with steam at high temperatures to produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, identify the limiting reactant. How many moles of each product are formed? 2.4 mol C G: W: R: 1 mol C = 1 mol H 2 mole H 2 G: amount of available limiting reactant H 2 is the first product 2.4 mol C G: W: R: 1 mol C = 1 mol CO mole CO G: amount of available limiting reactant CO is the second product Balanced Equation: C + H 2 O  H 2 + CO 2.4 mol C1 mol H 2 1 mol C = 2.4 mol H mol C 1 mol C 1 mol CO = 2.4 mol CO

80 Set VI: Reactions 2) Zn + Pb(NO 3 ) 2  Pb + Zn(NO 3 ) 2 3) Fe + 2HCl  H 2 + FeCl 2

81 III. Theoretical yield, Actual yield and Percent yield

82 C. Theoretical yield, Actual yield and Percent yield The yield of a chemical reaction is the quantity of product one obtains from a given ratio of reactants. The actual yield of a chemical reaction is the mass of the compound that you actually recover when you are done with the reaction. The actual yield is also referred to as the experimental yield The theoretical yield is the mass of compound you should obtain (theoretically ) if everything goes perfectly. In all of the examples above we have been pretending that everything is perfect. All of our wanteds have been theoretical.

83 Ex 1: What is the theoretical yield of Na 2 SO 4 in grams if 35 moles of NaOH is reacted with sufficient H 2 SO 4 ? Yield means product Theoretical yield means mathematically, what should you get? (this is why we do stoichiometry calculations)

84 Ex 1: What is the theoretical yield of Na 2 SO 4 in grams if 3.50 moles of NaOH is reacted with sufficient H 2 SO 4 ? = g Na 2 SO mol NaOH 2 mol NaOH mol Na 2 SO 4 x g Na 2 SO 4 x Given over 1Mole ratio1 mol = g 2NaOH + H 2 SO 4  Na 2 SO 4 + 2H 2 O G: W: R: 3.50 mol NaOH ? g Na 2 SO 4 2 mol NaOH = 1 mol Na 2 SO 4 1 mol Na 2 SO 4 = g Na 2 SO 4 2 different substances You see g Na 2 SO 4

85 If they give you the actual yield and you figure out the theoretical yield, you can find the percent yield. Actual Yield Theoretical Yield X 100= % Yield Actual Yield Theoretical Yield = % Yield 100 Same as…

86 Your percent yield should not be greater than 100 because the theoretical yield is the MAXIMUM yield you can have.

87 Example: From the examples above, if 221 grams of sodium sulfate were actually collected, what is the percent yield? 221 g NaSO g NaSO 4 X 100= 88.8% From the experiment From the calculation

88 Practice 1.A student conducts a single displacement reaction that produces 2.75 grams of copper. Mathematically he determines that grams of copper should have been produced. Calculate the student's percentage yield. 2.A student completely reacts 5.00 g of magnesium with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield? (*Hint: you must first find the theoretical yield.) 87.3% 97.7%

89 Practice 3. Consider the reaction below of zinc and hydrochloric acid to produce zinc chloride and hydrogen gas. Zn + 2HCl  H 2 +ZnCl 2 If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl 2 ? What is the actual yield in grams if the percent yield is 87%. Actual yield: 93.7 g

90 Practice 1 1.A student conducts a single displacement reaction that produces 2.75 grams of copper. Mathematically he determines that 3.15 grams of copper should have been produced. Calculate the student's percentage yield g Cu 3.15 g Cu Actual from the experiment calculated X 100 = 87.3 %

91 Practice 2 A student completely reacts 5.00g of magnesium with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield? (*Hint: you must first find the theoretical yield.) Actual yield Theo. X 100 = % yield actual

92 To calculate the theoretical yield Theoretical yield= g MgO g Mg g Mg mol Mg x 2 mol Mg 2 mol MgO x 2Mg + O 2  2 MgO G: W: R: 5.00 g Mg g MgO 2 mol Mg = 2 mol MgO 1 mol MgO = g MgO 2 different substances You see g Mg and g MgO 1 mol Mg = g Mg 1 mol MgO g MgO x

93 Practice 2 A student completely reacts 5.00g of magnesium with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield? (*Hint: you must first find the theoretical yield.) Actual yield 8.29 g MgO X 100= 97.7% 8.10 g MgO

94 Practice 3 3. Consider the reaction below of zinc and hydrochloric acid to produce zinc chloride and hydrogen gas. Zn + 2HCl  H 2 +ZnCl 2 If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl 2 ? What is the actual yield in grams if the percent yield is 87%. Actual yield: 93.7 g

95 If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl 2 ? What is the actual yield in grams if the percent yield is 87%. Theoretical yield= g ZnCl g HCl g HCl mol HCl x 2 mol HCl 1 mol ZnCl 2 x Zn + 2HCl  H 2 +ZnCl 2 G: W: R: 58.0 g HCl g ZnCl 2 2 mol HCl = 1 mol ZnCl 2 1 mol ZnCl 2 = g ZnCl 2 2 different substances You see g HCl & g ZnCl 2 1 mol HCl = g HCl 1 mol ZnCl g ZnCl 2 x

96 actual Practice 3 If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl 2 ? What is the actual yield in grams if the percent yield is 87% g 108 g Actual from the experiment calculated X 100 = 87 %


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