1 / 55

3 ways of representing the reaction of H 2 with O 2 to form H 2 O

3 ways of representing the reaction of H 2 with O 2 to form H 2 O. reactants. products. Chemical Equations. C 2 H 6 + O 2. CO 2 + H 2 O. NOT. 2C 2 H 6. C 4 H 12. Balancing Chemical Equations.

dacian
Download Presentation

3 ways of representing the reaction of H 2 with O 2 to form H 2 O

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 3 ways of representing the reaction of H2 with O2 to form H2O reactants products Chemical Equations

  2. C2H6 + O2 CO2 + H2O NOT 2C2H6 C4H12 Balancing Chemical Equations • Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water • Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.

  3. 1 carbon on right 6 hydrogen on left 2 hydrogen on right 2 carbon on left C2H6 + O2 C2H6 + O2 C2H6 + O2 CO2 + H2O 2CO2 + H2O 2CO2 + 3H2O Balancing Chemical Equations • Start by balancing those elements that appear in only one reactant and one product. start with C or H but not O multiply CO2 by 2 multiply H2O by 3

  4. multiply O2 by 4 oxygen (2x2) + 3 oxygen (3x1) 2 oxygen on left C2H6 + O2 2CO2 + 3H2O C2H6 + O2 2CO2 + 3H2O 2C2H6 + 7O2 4CO2 + 6H2O 7 7 2 2 Balancing Chemical Equations • Balance those elements that appear in two or more reactants or products. = 7 oxygen on right remove fraction multiply both sides by 2

  5. 4 C (2 x 2) 4 C Reactants Products 12 H (2 x 6) 12 H (6 x 2) 4 C 4 C 14 O (7 x 2) 14 O (4 x 2 + 6) 12 H 12 H 14 O 14 O 2C2H6 + 7O2 4CO2 + 6H2O Balancing Chemical Equations • Check to make sure that you have the same number of each type of atom on both sides of the equation.

  6. IS NOT How to “Read” Chemical Equations 2 Mg + O2 2 MgO 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO 2 grams Mg + 1 gram O2 makes 2 g MgO

  7. Chemical Equation Calculations Mass Atoms (Molecules) Molecular Weight Avogadro’s Number g/mol 6.02 x 1023 Molecules Reactants Products Moles

  8. Stoichiometry (more working with ratios) Ratios are found within a chemical equation. 2HCl + Ba(OH)2 2H2O + BaCl2 1 1 coefficients give MOLAR RATIOS 2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O and 1 mole of BaCl2

  9. Mole – Mole Conversions When N2O5 is heated, it decomposes: 2N2O5(g) 4NO2(g) + O2(g) a. How many moles of NO2 can be produced from 4.3 moles of N2O5? 2N2O5(g) 4NO2(g) + O2(g) ? mol 4.3 mol Units match 4.3 mol N2O5 = moles NO2 8.6 b. How many moles of O2 can be produced from 4.3 moles of N2O5? 2N2O5(g) 4NO2(g) + O2(g) 4.3 mol ? mol 4.3 mol N2O5 = mole O2 2.2

  10. gram ↔ mole and gram ↔ gram conversions When N2O5 is heated, it decomposes: 2N2O5(g) 4NO2(g) + O2(g) a. How many moles of N2O5 were used if 210g of NO2 were produced? 2N2O5(g) 4NO2(g) + O2(g) ? moles 210g Units match 210 g NO2 = moles N2O5 2.28 b. How many grams of N2O5 are needed to produce 75.0 grams of O2? 2N2O5(g) 4NO2(g) + O2(g) 75.0 g ? grams 75.0 g O2 = grams N2O5 506

  11. Gram to Gram Conversions Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl(aq)  AlCl3(aq) + H2(g) 2 6 2 3 First write a balanced equation.

  12. Gram to Gram Conversions Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl(aq)  AlCl3(aq) + H2(g) 2 6 2 3 ? grams 3.45 g Now let’s get organized. Write the information below the substances.

  13. gram to gram conversions Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl(aq)  AlCl3(aq) + H2(g) 2 6 2 3 ? grams 3.45 g Units match 3.45 g Al = g AlCl3 17.0 Let’s work the problem. We must always convert to moles. Now use the molar ratio. Now use the molar mass to convert to grams.

  14. 2CH3OH + 3O2 2CO2 + 4H2O grams CH3OH moles CH3OH moles H2O grams H2O 4 mol H2O 18.0 g H2O 1 mol CH3OH x = x x 2 mol CH3OH 32.0 g CH3OH 1 mol H2O Methanol burns in air according to the equation If 209 g of methanol are used up in the combustion, what mass of water is produced? molar mass CH3OH molar mass H2O coefficients chemical equation 209 g CH3OH 235 g H2O 3.8

  15. Problem: Calculating Reactants and Products in a Chemical Reaction Problem: Given the following chemical reaction between aluminum sulfide and water, if we are given 65.80 g of Al2S3: a) How many moles of water are required for the reaction? b) What mass of H2S & Al(OH)3 would be formed? Al2S3 (s) + 6 H2O (l) 2 Al(OH)3 (s) + 3 H2S (g) Plan: Calculate moles of aluminum sulfide using its molar mass, then from the equation, calculate the moles of water, and then the moles of hydrogen sulfide, and finally the mass of hydrogen sulfide using it’s molecular weight. Solution: a) molar mass of aluminum sulfide = moles Al2S3 =

  16. Calculating Reactants and Products in a Chemical Reaction a) cont. H2O: 0.4382 moles Al2S3 x b)H2S: 0.4382 moles Al2S3 x ___ moles H2O 1 mole Al2S3 ___ moles H2S 1 mole Al2S3 molar mass of H2S = mass H2S = Al(OH)3: 0.4382 moles Al2S3 x molar mass of Al(OH)3 = mass Al(OH)3 =

  17. Balanced reaction! Defines stoichiometric ratios! Unbalanced (i.e., non-stoichiometric) mixture! Limited by syrup!

  18. 6 green used up Limiting Reagents 1 green + 1 red  1 green-red 6 red left over 3.9

  19. Limiting Reactant In a chemical reaction where arbitrary amounts of reactants are mixed and allowed to react, the one that is used up first is the limiting reactant. A portion of the other reactants remains. There is a systematic procedure for finding the limiting reagent based on the reactant ratio (RR) defined as the ratio of the number of moles of a reactant to its coefficient in a balanced chemical equation. The reagent with the smallest reactant ratio is the limiting reactant.

  20. Limiting Reagent Concept

  21. Limiting Reagent Problems • Balance chemical equation • Determine limiting reagent • Do two separate calculations for the amount of product each reactant would produce if they were the limiting reagent • The reactant that gives the lower number is the limiting reagent. • The amount of product produced is the number calculated by limiting reagent

  22. Limiting Reagent Problems (cont) • Determination of the amount of excess reagent left over • Calculate the amount of excess reagent (ER) used in chemical reaction • Subtract the ER used from original amount of ER.

  23. Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) First copy down the the BALANCED equation! Now place numerical the information below the compounds.

  24. Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol ? moles 0.10 mol Hide one Two starting amounts? Where do we start?

  25. Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol ? moles 0.10 mol Hide Based on: KO2 0.15 mol KO2 = mol O2 0.1125

  26. Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol ? moles Hide 0.10 mol Based on: KO2 0.15 mol KO2 = mol O2 0.1125 0.10 mol H2O Based on: H2O = mol O2 0.150

  27. Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol ? moles 0.10 mol Based on: KO2 0.15 mol KO2 = mol O2 0.1125 It was limited by the amount of KO2. 0.10 mol H2O Based on: H2O = mol O2 0.150 H2O = excess (XS) reactant! What is the theoretical yield? Hint: Which is the smallest amount? The is based upon the limiting reactant?

  28. Do You Understand Limiting Reagents? 2Al + Fe2O3 Al2O3 + 2Fe g Al mol Al mol Fe2O3 needed g Fe2O3 needed OR g Fe2O3 mol Fe2O3 mol Al needed g Al needed 1 mol Fe2O3 160. g Fe2O3 1 mol Al = x x x 27.0 g Al 2 mol Al 1 mol Fe2O3 Start with 124 g Al need 367 g Fe2O3 In one process, 124 g of Al are reacted with 601 g of Fe2O3 Calculate the mass of Al2O3 formed. 367 g Fe2O3 124 g Al Have more Fe2O3 (601 g) so Al is limiting reagent 3.9

  29. 2Al + Fe2O3 Al2O3 + 2Fe g Al mol Al mol Al2O3 g Al2O3 1 mol Al x 27.0 g Al 1 mol Al2O3 102. g Al2O3 = x x 2 mol Al 1 mol Al2O3 Use limiting reagent (Al) to calculate amount of product that can be formed. 234 g Al2O3 124 g Al 3.9

  30. Actual Yield (mass or moles) Theoretical Yield (mass or moles) % Yield = x 100% Chemical Reactions in Practice: Theoretical, Actual, and Percent Yields Theoretical yield: The amount of product indicated by the stoichiometrically equivalent molar ratio in the balanced equation. Side Reactions: These form different products that take away from the theoretical yield of the main product. Actual yield: The amount of product that is actually obtained. Percent yield (%Yield): or Actual yield = Theoretical yield x (% Yield / 100%)

  31. Problem: Percent Yield Actual Yield Theoretical Yield Percent Yield = x 100% = Problem: The chemical reaction between iron and water to form the iron oxide, Fe3O4 and hydrogen gas is given below. If 4.55 g of iron is reacted with sufficient water to react all of the iron to form rust, what is the percent yield if only 6.02 g of the oxide are formed? Plan: Calculate the theoretical yield and use it to calculate the percent yield, using the actual yield. Solution: 3 Fe(s) + 4 H2O(l) Fe3O4 (s) + 4 H2 (g) Moles Fe = Theoretical moles Fe3O4 = Theoretical mass Fe3O4 =

  32. Problem: Percent Yield / Limiting Reactant Problem: Ammonia is produced by the Haber process using nitrogen and hydrogen Gas. If 85.90g of nitrogen are reacted with 21.66 g hydrogen and the reaction yielded 98.67 g of ammonia what was the percent yield of the reaction. N2 (g) + 3 H2 (g) 2 NH3 (g) Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reactant then calculate the theoretical yield, and then the percent yield. Solution: moles N2 = moles H2 =

  33. Problem: Percent Yield / Limiting Reactant N2 (g) + 3 H2 (g) 2 NH3 (g) Solution Cont. We have 3.066 moles of Nitrogen, and it is limiting, therefore the theoretical yield of ammonia is: mol NH3 = mass NH3 = Actual Yield Theoretical Yield Percent Yield = x 100% 98.67 g NH3 g NH3 Percent Yield = x 100% =

  34. Answers to Problems in the Lecture • (a) 2.629 moles H2O; (b) 44.81g H2S, 68.36g Al(OH)3 • 77.61 g Ca3(PO4)2 • 0.183 mol of AlCl3 • 30.0 g NO • 95.6 % • 94.49 %

  35. Limiting/Excess Reactant Problem with % Yield 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced? 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) ? g 120.0 g 47.0 g Hide one 120.0 g KO2 Based on: KO2 = g O2 40.51

  36. Limiting/Excess Reactant Problem with % Yield 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced? 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) ? g 120.0 g 47.0 g Hide 120.0 g KO2 Based on: KO2 = g O2 40.51 47.0 g H2O Based on: H2O = g O2 125.3 Question if only 35.2 g of O2 were recovered, what was the percent yield?

  37. If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced? 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) ? g 120.0 g 47.0 g 120.0 g KO2 Based on: KO2 = g O2 40.51 47.0 g H2O Based on: H2O = g O2 125.3 Determine how many grams of Water were left over. The Difference between the above amounts is directly RELATED to the XS H2O. 125.3 - 40.51 = 84.79 g of O2 that could have been formed from the XS water. 84.79 g O2 31.83 = g XS H2O

  38. The decomposition of KClO3 is commonly used to prepare small amounts of O2 in the laboratory: How many grams of O2 can be prepared from 4.50 g of KClO3?

  39. Propane, C3H8, is a common fuel used for cooking and home heating. What mass of O2 is consumed in the combustion of 1.00 g of propane?

  40. The most important commercial process for converting N2 from the air into nitrogen-containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3): How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2?

  41. Consider the reaction A mixture of 1.50 mol of Al and 3.00 mol of Cl2 is allowed to react. (a) Which is the limiting reactant? (b) How many moles of AlCl3 are formed? (c) How many moles of the excess reactant remain at the end of the reaction?

  42. A strip of zinc metal having a mass of 2.00 g is placed in an aqueous solution containing 2.50 g of silver nitrate, causing the following reaction to occur: (a) Which reactant is limiting? (b) How many grams of Ag will form? (c) How many grams of Zn(NO3)2 will form? (d) How many grams of the excess reactant will be left at the end of the reaction?

  43. Molarity Molarity is a term used to express concentration. The units of molarity are moles per liter (it is abbreviated as a capital M). When working problems, it is a good idea to change M into its units.

  44. Solutions A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution. What type of problem(s) is this? Molarity followed by dilution.

  45. Solutions A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution. 1st: 3.73 g = mol L 0.140 200.0 x 10-3 L molar mass of AlCl3 dilution formula M1V1 = M2V2 2nd: (0.140 M)(10.0 mL) = (? M)(100.0 mL) final concentration 0.0140 M = M2

  46. Solution Stoichiometry 50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used? H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g)

  47. Solution Stoichiometry 50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used? H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g) 50.0 mL ? g Our Goal 6.0 M = Look! A conversion factor!

  48. Solution Stoichiometry 50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used? H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g) 50.0 mL ? g Our Goal 6.0 M = NaHCO3 2 mol H2SO4 50.0 mL NaHCO3 84.0 g 50.4 = g NaHCO3 mol NaHCO3 1 mol H2SO4

More Related