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Unit 9: Stoichiometry

Unit 9: Stoichiometry. Stoichiometry. The process of using a balanced chemical equation to determine the relative amounts of reactants and products involved in a reaction. What do the coefficients tell us?. 1 CO (g) + 2 H 2(g)  1 CH 3 OH (l)

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Unit 9: Stoichiometry

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  1. Unit 9: Stoichiometry

  2. Stoichiometry • The process of using a balanced chemical equation to determine the relative amounts of reactants and products involved in a reaction.

  3. What do the coefficients tell us? 1 CO(g) + 2 H2(g) 1 CH3OH(l) The coefficients can be used to describe the reaction The coefficients show relative amounts of products and reactants The coefficients can have units of moles or molecules 1 molecule CO + 6.02 x 1023 CO molecules +1 mol CO molecules + 2 molecules H2 1 molecule CH3OH 2 (6.02 x 1023) H2 molecules  6.02 x 1023 CH3OH molecules 2 molH2 molecules  1 molCH3OH molecules

  4. Balancing Equations • The first step to all STOICHIOMETRY problems is to balance the equation. • Balance the following equation: ___C3H8 (g) + ___ O2 (g)___CO2 (g) + ___ H2O (l) C H O

  5. Mole Ratio • Mole ratio= the ratio of moles of one substance to moles of another substance in a balanced chemical equation • Ex: 2Na + Cl2 2NaCl • What is the mole ratio of sodium to chlorine gas? • What is the mole ratio of sodium to sodium chloride? • What is the mole ratio of chlorine to sodium chloride? • Why is a BALANCED chemical equation important? • In stoichiometry mole ratios are used as conversion factors. • Answers to all stoichiometry problems should be in correct significant figures and have units followed by a substance (ex: 4.0 moles of NaCl)

  6. Moleto Mole Relationships • We can used a balanced chemical equation to predict the amount of products that a given number of moles of reactant will yield or amount of reactant needed for a certain amount of product • Moles are used instead of mass because each element/compound has a different mass • A balanced equation is needed to make a mole to mole comparison • The coefficients in a balanced equation will give us the mole to mole ratio • Mole ratios are used as conversion factors

  7. Mole to Mole Ex #1 • Problem: In a synthesis reaction between nitrogen and hydrogen, how many moles of hydrogen is needed to make 4.0 moles of ammonia? • First Step: Write a balanced chemical equation: N2(g) + 3 H2(g)  2 NH3(l) • Second Step: Determine mole ratio needed (coefficients): For every 3 moles of H2(g)used, it produces2 moles of NH3(l) • Third Step: Set up a T chart with conversion factors ALWAYS START WITH YOUR GIVEN: (given 4.0 moles of ammonia) x = 6.0 moles of H2

  8. Mole to Mole Ex #2 • Problem: In a synthesis reaction between nitrogen and hydrogen, how many moles of nitrogen are needed to react with 6.0 moles of hydrogen? • First Step: Write a balanced equation: N2(g) + 3 H2(g)  2 NH3(l) • Second Step: Determine mole ratio (coefficients): For every 3 moles of H2(g)used, we need 1 moles of N2(g) • Third Step: Set up a T chart with conversion factors ALWAYS START WITH YOUR GIVEN: (given 6.0 moles of hydrogen) x = 2.0 moles of N2

  9. Mole to Mole Ex #3 • For a synthesis reaction between hydrogen and oxygen to make water answer the following: • A) How many moles of H2O are produced when 5.00 moles of oxygen are used? • B) If 3.00 moles of H2O are produced, how many moles of oxygen are needed? • C) How many moles of hydrogen gas are needed to produce 4 moles of water? • A) 10.0 moles of H2O • B) 1.50 moles of O2 • C) 4 moles of H2

  10. Converting Using the Mole • Mass to Moles: molar mass (g) • Volume to Moles: 22.4L • Particles to Moles: 6.02 x 1023 particles • Moles of Substance A to Moles of Substance B: Mole ratio • Must be in units of moles to use the mole ratio!

  11. Mole: The Go Between Volume of gas 1 mole 22.4 L 22.4 L 1 mole MOLE 1mol 6.02 x 1023 particles # of particles Molar mass (g) 1 mol 6.02 x 1023 particles 1mol Mass 1 mol Molar mass (g)

  12. Stoichiometry • When performing stoichiometric conversions you may begin in units of: mass, volume, particles, or moles • You will be asked to convert to units of: mass, volume, particles, or moles • Always start with a balanced chemical equation • Always convert to moles and then use the mole ratio • Always start with the value given in the problem • Include sig figs, units, and a substance in your answer • Always show your work

  13. Stoichiometry Ex #4 • Problem: Oxygen gas can be produced by decomposing potassium chlorate (potassium chloride is also produced). If 138.6 g of KClO3 is heated and decomposes completely, what mass of oxygen gas is produced? • How do we solve? • Always start with a balanced chemical equation • We are given the # grams of potassium chlorate, so we must convert to moles of potassium chlorate because the balanced equation deals in moles rather than grams • Next we can use the coefficients in the balanced equation to determine the mole ratio • Finally, we will use the molar mass of oxygen gas to calculate grams of oxygen gas produced

  14. Mass to Mass Ex #4 • Write a balanced chemical equation: 2  KClO3 (s)  --> 2 KCl (s)  + 3 O2 (g)           • Find the molar mass of KClO3 and O2: KClO3= 122.55g/mol O2 = 32.00 g/mol • Find mole ratio: 2 moles of KClO3reacts to produce 3 moles of O2 • Use a T chart, start with given value: x x = 54.29g of O2 is produced

  15. Mass to Volume Ex #5 • A chemical reaction between diboronhexahydride and oxygen gas will produce Oxoborinicacid (HBO2)and water. A.) What volume, at STP, of O2 will be needed to burn 36.1g of diboronhexahydride? B.) How many moles of water are produced from burning B2H6? • What is always the first step? • What is given? • What do we want to determine? (there are two questions in one) • What information do we need to know? • What is the mole ratio?

  16. Answer: Practice Problem • Balance the equation: B2H6 + 3 O2  2 HBO2 + 2 H2O • Fine the mole ratio • Find Molar Mass needed • Write T Chart with conversion factors: A. xxx=87.7 L O2 B. xx = 2.61 moles H2O

  17. S’mores • What is the “formula” for making s’mores? • How many s’mores could I make if I had 12 gram crackers? • How many s’mores could I make if I had 12 marshmallows? • How many s’mores could I make if I had 12 chocolate squares? • If I have 12 gram crackers, 12 marshmallows, and 12 chocolate squares, how many s’mores could I make? • Which ingredient limits the amount of s’mores that can be made? • Which ingredients are excess (left over?)? How much excess of those ingredients would you have?

  18. Limiting Reactant v. Excess Reactant The Excess Reactant does not limit or stop a reaction from occurring(also called excess reagent) There will be an excess or left over of this reactant. The Limiting Reactantis the reactant which limits how much the product is produced (also called limiting reagent). The limiting reactant is completely used and none is left over. The Limiting Reactant stops or limits the reaction after it is entirely consumed. The limiting reactant determines how much product is made

  19. Limiting Reactant • To determine which reactant is the limiting reactant • 1st Convert both reactants to moles or grams of product. • 2nd Compare the amounts of product that can be produced. • The reactant that produces the least amount of product is the limiting reactant • The reactant that produces the most product is the excess reactant.

  20. Limiting Reagent Ex #6 Problem: Aluminum reacts with chlorine gas to form aluminum chloride.If you have 34.0 g of Al and 39.0 g of Cl2, what is the limiting reactant? What is the excess reactant? How many moles of aluminum chloride are produced? • Determine the limiting reagent by finding the # of mols of product each reactant can possibly produce: Al : x x = 1.26 mol AlCl3 Cl2: x x = 0.367 molAlCl3 • Which ever reactant creates less product is the limiting reactant, the other reactant is the excess reactant. Chlorine gas is the limiting reactant and Aluminum is the excess reactant • Determining the limiting reactant also determines how much product can be made: 0.367 molAlCl3

  21. Limiting Reagent Ex #7 Problem: When 3.22 moles of Al reacts with 4.96 moles of HBr, how many moles of H2 are formed, considering the reaction below? What is the limiting reactant and the excess reactant? How much excess reactant remains (hint: first determine how much was used)? 2 Al + 6 HBr → 2 AlBr3 + 3 H2 Part 1: • Al: = 4.83 molH2 • HBr: = 2.48 mol H2 • 2.48 mol H2 are produced HBr is the limiting reactant and Al is the excess reactant

  22. Limiting Reagent Ex #7 Problem: When 3.22 moles of Al reacts with 4.96 moles of HBr, how many moles of H2 are formed, considering the reaction below? What is the limiting reactant and the excess reactant? How much excess reactant remains (hint: first determine how much was used)? 2 Al + 6 HBr → 2 AlBr3 + 3 H2 Part 2: • = 1.65 mol Al USED • Started with 3.22 moles of Al and USED 1.65 mol Al • 3.22 moles Al – 1.65 mol Al = 1.57 mol Al excess

  23. True or False? • The limiting reactant is ALWAYS the reactant that is present in the lesser amount. • The limiting reactant ALWAYS determines how much product can be made. • There can be more than one limiting reactant. • There is ALWAYS a limiting reactant.

  24. Limiting Reagent Practice • Ex #8: How many grams of aluminum sulfate are produced if 23.33 g Al reacts with 74.44 g CuSO4? • How much excess reactant remains?    Al (s)  +  CuSO4(aq)   Al2(SO4)3(aq)  +  Cu (s)

  25. ANSWER • 2Al (s)  +  3CuSO4(aq)   Al2(SO4)3(aq)  +  3Cu (s) • Limited reactant is CuSO4 and it produces 53.19 g of Al2(SO4)3 • Only 8.39 g of Al is needed to react with 74.44 grams of CuSO4 • 23.33- 8.39= 14.9 grams of Al is left over

  26. Theoretical Yield is how much product SHOULD be produced in ideal conditions. This number is calculated using stoichiometry The theoretical yield is determined by finding the limiting reagent and determining how much product should be made Actual Yield is how much product was ACTUALLY synthesized in an experiment. This number is an experimental value. Percent Yield: is the ratio between the actual yield and the theoretical yield multiplied by 100%. It indicates efficiency. Theoretical, Actual, and Percent Yield Percentage Yield =     Mass of Actual Yield       x   100%                   Mass of Theoretical Yield

  27. 1 mol H2 actual 2 mol H2O 138 g H2O x = x 2.02 g H2 theoretical 143 g H2O 2 mol H2 18.02 g H2O x 1 mol H2O = = 143 g 96.7%- Very efficient Theoretical Yield Example • Ex #9: What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2? Is this efficient? Step 1: Write the balanced chemical equation 2H2 + O2 2H2O Step 2: Determine actual and theoretical yield (actual is given, theoretical is calculated): # g H2O= 16 g H2 Step 3: Calculate % yield % yield = x 100% x 100%

  28. actual 2 mol NH3 40.5 g NH3 x = theoretical 3 mol H2 227 g NH3 17.04 g NH3 x 1 mol NH3 = = 227 g 17.8% Not efficient Theoretical Yield Practice Ex #10: What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2? Is this process efficient? Step 1: write the balanced chemical equation N2+ 3H2 2NH3 Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: # g NH3= 20.0 mol H2 Step 3: Calculate % yield % yield = x 100% x 100%

  29. 1 mol O2 actual 1 mol H2 58 g H2O 2 mol H2O 2 mol H2O x x x x = 2.02 g H2 32 g O2 theoretical 62.4 g H2O 1 mol O2 2 mol H2 18.02 g H2O 18.02 g H2O x x 1 mol H2O 1 mol H2O 92.9% Very Efficient = = 68 g H2O = 62.4 g H2O Example #11 Ex #11: What is the % yield of H2O if 58 g H2O are produced by combining 60. g O2 and 7.0 g H2?Is this process efficient? How much excess reactant remains? 60 g O2 7.0 g H2 % yield = x 100% x 100%

  30. 1 mol H2 1molO2 x x 2.02 g H2 2 mol H2 32.00 g O2 x 1 molO2 = 55 g O2USED Example #11 Ex #11: What is the % yield of H2O if 58 g H2O are produced by combining 60. g O2 and 7.0 g H2?Is this process efficient? How much excess reactant remains? 7.0 g H2 58 g – 55 g = 3 g O2 remain

  31. More Percent Yield Questions • Ex #12: The electrolysis of water forms H2 and O2. 2H2O  2H2 + O2 What is the % yield of O2 if 12.3 g of O2 is produced from the decomposition of 14.0 g H2O? • Ex #13: 107 g of oxygen is produced by heating 300g of potassium chlorate. Calculate % yield. 2KClO3 2KCI + 3O2 • Ex #14: What is the % yield of iron (II) sulfide if 3.00 moles of Fe reacts with excess sulfur to produce 220 grams of ferrous sulfide?Fe + S FeS

  32. 1 mol H2O actual 1 mol O2 12.3 g O2 x = x 18.02 g H2O theoretical 12.43 g O2 2 mol H2O 32 g O2 x 1 mol O2 = = 12.43 g 98.9% Example 12 The electrolysis of water forms H2 & O2. 2H2O  2H2 + O2 Give the percent yield of O2 if 12.3 g O2 is produced from the decomp. of 14 g H2O? • Actual yield is given: 12.3 g O2 • Next, calculate theoretical yield # g O2= 14.0 g H2O Finally, calculate % yield % yield = x 100% x 100%

  33. 1 mol KClO3 actual 3 mol O2 107 g O2 x = x 122.55 g KClO3 theoretical 117.5 g O2 2 mol KClO3 32 g O2 x 1 mol O2 = = 117.5 g 91.1% Example 13 107 g of oxygen is produced by heating 300 grams of potassium chlorate. 2KClO3 2KCI + 3O2 • Actual yield is given: 107 g O2 • Next, calculate theoretical yield # g O2= 300 g KClO3 Finally, calculate % yield % yield = x 100% x 100%

  34. actual 1 mol FeS 220 g O2 x = theoretical 1 mol Fe 263.7 g O2 87.91 g FeS x 1 mol FeS = = 83.4% 263.7 g Example 14 What is % yield of ferrous sulfide if 3 mol Fe produce 220 grams of ferrous sulfide? Fe + S FeS • Actual yield is given: 220 g FeS • Next, calculate theoretical yield # g FeS= 3.00 mol Fe Finally, calculate % yield % yield = x 100% x 100%

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