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

III. Stoichiometry Stoy – kee – ahm –eh - tree. Chapter 12. 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. Things you should remember.

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

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  1. III. StoichiometryStoy – kee – ahm –eh - tree Chapter 12

  2. SectionsClick 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 1023 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 • stoikheion, meaning element • metron, meaning measure • Thus Stoichiometry- measuring elements!

  5. III. Stoichiometry • Stoichiometry (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 2 __H2SO4(aq) + __NaHCO3(s)  __Na2SO4(aq) + __H2O(l) + __CO2(g) 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. For the Reaction:H2SO4 + 2NaHCO3Na2SO4 + 2H2O + 2CO2 What’s the relationship between NaHCO3 and Na2SO4? 2 mole NaHCO3 = 1 mol Na2SO4 (it’s just the coefficients!!) What’s the relationship between H2SO4 and CO2? 1 mol H2SO4 = 2 mol CO2 (it’s just the coefficients!!)

  11. For the Reaction:H2SO4 + 2NaHCO3Na2SO4 + 2H2O + 2CO2 • 1 mol H2SO4 = 2 mol NaHCO3 or • 1 mc H2SO4 = 2 mc NaHCO3 or • 1 mol/mc H2SO4 2 mol/mc NaHCO3 or • 2 mol/mc NaHCO3 1 mole/mc H2SO4

  12. For the Reaction:H2SO4 + 2NaHCO3Na2SO4 + 2H2O + 2CO2 • 1 mol H2SO4 = 2 mol CO2 or • 1 mc H2SO4 = 2 mc CO2 or • 1 mol/mc H2SO4 2 mol/mc CO2 or • 2 mol/mc CO2 1 mole/mc H2SO4

  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.) Fe + O2 Fe2O3 Skeletal equation: 4Fe + 3O2 2Fe2O3 Balanced equation: 4 mol Fe = 3 mol O2 Relationships: 4 mol Fe = 2 mol Fe2O3 3 mol O2 = 2 mol Fe2O3

  14. Practice 1. Write three mole ratios (relationships) from the reaction below: Al2S3 + H2O  Al(OH)3 + H2S 6 2 3 1 mol Al2S3 = 6 mol H2O 1 mol Al2S3 = 2 mol Al(OH)3 1 mol Al2S3 = 3 mol H2S You should have 3 6 mol H2O = 2 mol Al(OH)3 6 mol H2O = 3 mol H2S 2 mol Al(OH)3 = 3 mol H2S

  15. Practice 1. Write three mole ratios (relationships) from the reaction below: Al2S3 + H2O  Al(OH)3 + H2S 6 2 3 1 mol Al2S3 = 6 mol H2O 1 mol Al2S3 = 2 mol Al(OH)3 1 mol Al2S3 = 3 mol H2S 6 mol H2O = 2 mol Al(OH)3 6 mol H2O = 3 mol H2S

  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. 2Al2O3  4Al + 3O2 2 mol Al2O3 = 4 mol Al 2 mol Al2O3 = 3 mol O2 4 mol Al = 3 mol O2

  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 1023of 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: • put your GIVEN OVER 1 • place your relationships where the units cancel out diagonally • everything equal to each other goes above and below each other • 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: N2 (g) + 3H2 (g)  2NH3 (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: N2 (g) + 3H2 (g)  2NH3 (g) No grams, No molar mass!! Given : Want : Relationships: 4.00 mol H2 No fun, mc, No 6.02 x 1023!! ? mol NH3 2 mol NH3 = 3 mol H2 4.00 mol H2 2 mol NH3 = mol NH3 2.67 x 3 mol H2 1

  26. Ex 2: How many moles of nitrogen were used if 7.8 moles of ammonia were made in excess hydrogen gas? Balanced Equation: N2 (g) + 3H2 (g)  2NH3 (g) No grams, No molar mass!! Given : Want : Relationships: 7.8 mol NH3 No fun, mc, No 6.02 x 1023!! ? mol N2 2 mol NH3 = 1 mol N2 7.8 mol NH3 1 mol N2 = mol N2 3.9 X 2 mol NH3 1

  27. Ex 3: How many moles of hydrogen react with 13 moles of nitrogen? Balanced Equation: N2 (g) + 3H2 (g)  2NH3 (g) No grams, No molar mass!! Given : Want : Relationships: 13 mol N2 No fun, mc, No 6.02 x 1023!! ? mol H2 1 mol N2 = 3 mol H2 13 mol N2 3 mol H2 = moles H2 39 x 1 mol N2 1

  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 1023of 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: N2 (g) + 3H2 (g)  2NH3 (g) How do I know when to use mole ratios, molar mass or 6.02 x 1023? Given : Want : Relationships: 4.00 mol H2 ? g NH3 2 mol NH3 = 3 mol H2 1 mol NH3 = 17.031g NH3 Look at the “given” and “want” for clues No fun, mc, No 6.02 x 1023!!

  31. Ex 1: How many grams of ammonia are produced from 4 moles of hydrogen gas in excess nitrogen? Balanced Equation: N2 (g) + 3H2 (g)  2NH3 (g) Given : Want : Relationships: 4.00 mol H2 ? g NH3 2 mol NH3 = 3 mol H2 1 mol NH3 = 17.031g NH3 4.00 mol H2 2 mol NH3 17.031g NH3 x x 3 mol H2 1 mol NH3 1 = g NH3 45.4

  32. Ex 2: How many moles of ammonia are produced from 4.00 grams of hydrogen gas in excess nitrogen? Balanced Equation: N2 (g) + 3H2 (g)  2NH3 (g) Given : Want : Relationships: 4.00 g H2 ? mol NH3 2 mol NH3 = 3 mol H2 1 mol H2 = 2.016g H2 Remember your Steps to Success! No fun, mc, No 6.02 x 1023!!

  33. Ex 2: How many moles of ammonia are produced from 4 grams of hydrogen gas in excess nitrogen? Balanced Equation: N2 (g) + 3H2 (g)  2NH3 (g) Given : Want : Relationships: 4.00 g H2 ? mol NH3 2 mol NH3 = 3 mol H2 1 mol H2 = 2.016g H2 4.00 g H2 1 mol H2 2 mol NH3 x x 2.016g H2 3 mol H2 1 = mol NH3 1.32

  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 N2 163 g H2 1.23 mol N2

  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: N2 (g) + 3H2 (g)  2NH3 (g) Given : Want : Relationships: 4.00 g H2 ? g NH3 2 mol NH3 = 3 mol H2 1 mol H2 = 2.016g H2 Remember your Steps to Success! 1 mol NH3 = 17.031 g NH3

  37. Ex 1: How many grams of ammonia are produced from 4 grams of hydrogen gas in excess Nitrogen? Balanced Equation: N2 (g) + 3H2 (g)  2NH3 (g) Given : Want : Relationships: 4.00 g H2 ? g NH3 2 mol NH3 = 3 mol H2 1 mol H2 = 2.016g H2 1 mol NH3 = 17.031g NH3 17.031g NH3 1 mol H2 2 mol NH3 4.00 g H2 x x x 2.016g H2 3 mol H2 1 mol NH3 1 = g NH3 22.5

  38. Ex 2: How many grams of Nitrogen react with 12.0 grams of Hydrogen? Balanced Equation: N2 (g) + 3H2 (g)  2NH3 (g) Given : Want : Relationships: 12.0 g H2 ? g N2 1 mol H2 = 2.016g H2 Remember your Steps to Success! 1 mol N2 = 28.014 g N2

  39. Ex 2: How many grams of Nitrogen react with 12.0 grams of Hydrogen? Balanced Equation: N2 (g) + 3H2 (g)  2NH3 (g) Given : Want : Relationships: 12.0 g H2 ? g N2 1 mol N2 = 3 mol H2 1 mol H2 = 2.016g H2 1 mol N2 = 28.014 g N2 28.014g N2 12.0 g H2 1 mol H2 1 mol N2 x x x 2.016g H2 1 3 mol H2 1 mol N2 = g N2 55.6

  40. Practice • Determine the mass of NH3 produced from 280 g of N2. • What mass of nitrogen is needed to produce 100. kg of ammonia? 340 g NH3 8.22 x 104 g N2

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

  42. Ex 1: How many grams of ammonia are produced from 2.09 x 1014 mc of hydrogen gas? Balanced Equation: N2 (g) + 3H2 (g)  2NH3 (g) Given : Want : Relationships: 2.09 x 1014 mc H2 ? g NH3 2 mol NH3 = 3 mol H2 1 mol NH3 = 17.031g NH3 1 mol H2 = 6.02 x 1023 mc H2

  43. Ex 1: How many grams of ammonia are produced from 2.09 x 1014 mc of hydrogen gas? Balanced Equation: N2 (g) + 3H2 (g)  2NH3 (g) Given : Want : Relationships: 2.09 x 1014 mc H2 ? g NH3 2 mol NH3 = 3 mol H2 1 mol H2 = 6.02 x 1023 mc H2 1 mol NH3 = 17.031g NH3 1 mol H2 2 mol NH3 2.09 x 1014 mc H2 17.031g NH3 x x x 3 mol H2 1 6.02 x 1023 mc H2 1 mol NH3 3.94 x 10-9 = g NH3

  44. Ex 2: How many molecules of ammonia are produced from 40.2 g of H2 in the presence of excess N2? Balanced Equation: N2 (g) + 3H2 (g)  2NH3 (g) Given : Want : Relationships: 40.2g H2 ? mc NH3 2 mol NH3 = 3 mol H2 1 mol H2 = 2.016g H2 1 mol NH3 = 6.02 x 1023 mc NH3

  45. Ex 2: How many molecules of ammonia are produced from 40.2 g of H2 in the presence of excess N2? Balanced Equation: N2 (g) + 3H2 (g)  2NH3 (g) Given : Want : Relationships: 40.2g H2 ? mc NH3 2 mol NH3 = 3 mol H2 1 mol H2 = 2.016g H2 1 mol NH3 = 6.02 x 1023 mc NH3 1 mol H2 40.2g H2 2 mol NH3 6.02 x 1023 mc NH3 x x x 2.016g H2 3 mol H2 1 1 mol NH3 8.00 x 1024 = mc NH3

  46. Practice • How many mc of nitrogen will react with exactly 7.04 grams of hydrogen? • How many mc of ammonia will 2.09 x 1021 mc of N2 produce in excess hydrogen? 7.01 x 1023 mc N2 4.18 x 1021 mc NH3

  47. 1) How many mc of nitrogen will react with exactly 7.04 grams of hydrogen? Balanced Equation: N2 (g) + 3H2 (g)  2NH3 (g) Given : Want : Relationships: 7.04 g H2 ? mc N2 1 mol N2 = 3 mol H2 1 mol H2 = 2.016g H2 1 mol N2 = 6.02 x 1023 mc N2 1 mol H2 7.04 g H2 1 mol N2 6.02 x 1023 mc N2 x x x 2.016g H2 3 mol H2 1 1 mol N2 7.01 x 1023 = mc N2

  48. How many mc of ammonia will 2.09 x 1021 mc of N2 produce in excess hydrogen? Balanced Equation: N2 (g) + 3H2 (g)  2NH3 (g) Given : Want : Relationships: 2.09 x 1013 mc N2 ? mc NH3 2 mol NH3 = 1 mol N2 1 mol N2 = 6.02 x 1023 mc N2 1 mol NH3 = 6.02 x 1023 NH3 2.09 x 1013 mc N2 1 mol N2 2 mol NH3 6.02 x 1023 mc NH3 x x x 1 1 mol N2 1 mol NH3 6.02 x 1023 mc N2 4.18 x 1021 = mc NH3

  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.

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