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Stoichiometry

Stoichiometry. What is stoichiometry?. Chemists are constantly faced with the question, "How much?" How much of an element is present in a compound? How much of each element or compound is required to produce a new compound?

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Stoichiometry

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  1. Stoichiometry

  2. What is stoichiometry? • Chemists are constantly faced with the question, "How much?" How much of an element is present in a compound? How much of each element or compound is required to produce a new compound? • Fortunately, all of these questions can be answered using the same procedure. Solving these types of questions is called STOICHIOMETRY.

  3. N2 (g) + 3 H2 (g) 2 NH3 (g) • A lot of information can be concluded from a balanced equation! • Particles: • 1 molecule of N2 • 3 molecules of H2 • 2 molecules of NH3 • 2 nitrogen atoms (reactants and products) • 6 hydrogen atoms (reactants and products) 1: 3: 2 ratio

  4. N2 (g) + 3 H2 (g) 2 NH3 (g) • Moles: (Avogadro’s number of representative particles is one mole of a substance.) • 1 mole of N2 • 3 mole of H2 • 2 mole of NH3 1: 3: 2 ratio • The amount of moles in the reactant will not always equal the amount of moles in the product

  5. N2 (g) + 3 H2 (g) 2 NH3 (g) • Mass • 1 mole N2: 2 (14.0) = 28.0 grams • 3 mole H2: 6 (1.0) = 6.0 grams • Total: 28.0 + 6.0 = 34.0 grams • 2 moles of NH3 = 34.0 grams The mass from the reactants will always equal the mass of the products

  6. N2 (g) + 3 H2 (g) 2 NH3 (g) • Volume • 1 mole N2: = 1mole x 22.4 L = 22.4 L • 3 mole H2: = 3 mole x 22.4 L = 67.2 L • 2 moles of NH3 = 2 mole x 22.4 L = 44.8 L * Volume of the reactant is NOT always equal to the volume of the products (there are different amount of moles)*

  7. Mole to Mole Calculations • N2 (g) + 3 H2 (g) 2 NH3 (g) • The coefficients from the balanced equation are used to establish mole ratios. • 1 mol N2 : 3 mol H2 : 2 mol NH3

  8. How many moles of ammonia are produced when 0.60 moles of nitrogen reacts with excess hydrogen?N2 (g) + 3 H2 (g) 2 NH3 (g) • Set up a proportion (make sure to cross multiple) • 0.60 mol N2= 1 mol N2 =1.2 mol NH3 x mol NH3 2 mol NH3

  9. Try This!4 Al + 3 O2 --> 2 Al2O3 • How many moles of aluminum are needed to form 3.7 moles of Al2O3? • 3.7 mol Al2O3=2 mol Al2O3= 7.4 mol Al x mol Al 4 mol Al • How many moles of oxygen molecules will be needed to produce 10. moles of Al2O3 ? • 10. mol Al2O3=2 mol Al2O3= 15 mol O2 x mol O2 3 mol O2

  10. Mass to Mass Calculations • Calculate the number of grams of NH3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen. The balanced equation is… • N2 (g) + 3 H2 (g)  2 NH3

  11. N2 (g) + 3 H2 (g)  2 NH3 • 5.40 g H2 = 6.0 g H2 x g NH3 34.0 g NH3 or • 5.40 g H2= 2.7 mol H2=3 mol H2= 1.8 mol NH3 x 17.0 g NH3 = 2.0 g H2 x mol NH3 2 mol NH3 31 g NH3

  12. Try This! • CaC2 + 2 H2O --> C2H2 + Ca(OH)2 • How many grams of C2H2 are produced by adding water to 5.00 grams of CaC2? • 5.00 g CaC2=64.1 g CaC2 x g C2H2 26.0 g C2H2 • 2.03 g C2H2

  13. Other Stoichiometry Questions • How many molecules of oxygen are produced when a sample of 29.2 g of water is decomposed by electrolysis according to this balanced equation? 2 H2O (l)  2 H2 (g) + O2 (g) (Look at p. 248 in your text)

  14. 2 H2O (l)  2 H2 (g) + O2 (g) • 29.2g H2O=1.62 mol H2O = 0.811 mol O2 x 6.02 x 1023 molecule O2 18.0 g H2O 2 = 4.88 x 10 23 molecules O2

  15. The “jist” of it… • In order to calculate stoichiometry problems, find your known and your unknown. Then solve for your unknown. • Given quantity  change to moles  mole ratio  change to wanted quantity or gram to gram proportion

  16. Methanol CH3OH is used in the production of many chemicals. Methanol is made by reacting carbon monoxide and hydrogen at high temperature and pressure. CO + 2H2 --> CH3OH • How many moles of each reactant are needed to produce 359 g CH3OH? • Calculate the volume of H2 needed to produce 4.00 moles of CH3OH. • How many molecules of hydrogen are necessary to react with 2.85 moles CO?

  17. How many moles of each reactant are needed to produce 359 g CH3OH? CO + 2H2 --> CH3OH 359 g CH3OH =11.21.. mol CH3OH =1 mol CH3OH= 11.2 mol CO 32.0 g CH3OH x mol CO 1 mol CO 359 g CH3OH =11.21.. mol CH3OH =1 mol CH3OH= 22.4 mol H2 32.0 g CH3OH x mol H2 2 mol H2

  18. Calculate the volume of H2 needed to produce 4.00 moles of CH3OH at STP.CO + 2H2 --> CH3OH • 4. 00 mol CH3OH=1 mol CH3OH = 8.00 mol x 22.4 L = 179.2 liters x mol H2 2 mol H2

  19. How many molecules of hydrogen are necessary to react with 2.85 moles of CO?CO + 2H2 --> CH3OH • 2.85 mol CO =1 mol CO = 5.70 mol H2 x 6.02 x 1023 molecules x mol H2 2 mol H2 = 3.43 x 1024 molecules

  20. Volume to Volume • 2 CO + O2 --> 2 CO2 • How many liters of oxygen are required to burn 3.86 L of carbon monoxide?

  21. Answer • 3.86 L CO =44.8 L CO x L O2 22.4 L O2 = 1.93 L O2

  22. Limiting/Excess Reagent • If you have 24 pieces of paper, 12 envelopes, and 10 stamps how many letters can you mail? • Only ten because you only have ten stamps • The limiting reagent is used up. • There is extra of the excess reagent(s).

  23. Sodium chloride can be prepared by the reaction of sodium metal with chlorine gas. 2 Na + Cl2 2 NaCl • Suppose that 6.70 moles Na react with 3.20 moles Cl2. • First, find the limiting and excess reagent (reactant). • Find how many moles are needed to carry out the reaction.

  24. 2 Na (s) + Cl2 (g)  2 NaCl • 6.70 mol Na=2 mol Na = 3.35 moles Cl2 x mol Cl2 1 mol Cl2 This means that 3.35 moles of Cl2 is needed to react with 6.70 moles of Na. Because only 3.20 moles of Cl2 are available, chlorine becomes the limiting reagent. Sodium is in excess.

  25. 2 Na (s) + Cl2 (g)  2 NaCl • 3.2 mol Cl2=1 mol Cl2= 6.4 moles Na x mol Na 2 mol Na This means that 6.4 moles of Na is needed to react with 3.2 moles of Cl2. Because 6.7 moles of Na are available, Sodium is in excess.

  26. 2 Na (s) + Cl2 (g)  2 NaCl • How many moles of NaCl are produced? • In order to answer the question….your given is the limiting reagent (3.20 moles Cl2). • Put the wanted (the product) as the variable. • 3.20 mol Cl2=1 mol Cl2= 6.40 mol NaCl x mol NaCl 2 mol NaCl

  27. Write the balanced equation for the reaction lead (II) nitrate with sodium iodide to form sodium nitrate and lead (II) iodide. • If you have 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?

  28. This problem can be solved two different ways…. • 1)Find how many grams is required of each reactant to completely react with the other reactant. Then use the limiting reactant as your given. • 2) Do a mass to mass stoichiometry problem using both of the reactants. The reactant that makes the smaller amount of product is your limiting reagent.

  29. Pb(NO3)2 + 2NaI --> PbI2 + 2 NaNO3 25 g Pb(NO3)2=331.2 g Pb(NO3)2 x g NaI 299.8 g NaI = 22. 6 g of NaI 22. 6 grams of NaI are needed to completely react with 25 grams of Pb(NO3)2. Since we only have 15 grams of NaI, it becomes our limiting reactant.

  30. Pb(NO3)2 + 2NaI --> PbI2 + 2 NaNO3 NaI is the limiting reagent! Then find the amount of product that can be formed from the amount of sodium iodide given. NaI = 2(23.0 + 126.9) = 299.8 g NaNO3 = 2((23.0 + 14.0 + 3(16.0)) = 170.0 g 15 g NaI = 299.8 g NaI = 8.5 g NaNO3 x g NaNO3 170.0 g NaNO3

  31. Another Way! Do a mass to mass stoichiometry problem using both of the reactants. The reactant that makes the smaller amount of product is your limiting reagent.

  32. Pb(NO3)2 + 2 NaI --> PbI2 + 2 NaNO3 15 g NaI = 299.8 g NaI = 8.5 g NaNO3 x g NaNO3 170.0 g NaNO3 25 g Pb(NO3)2= 331.2 g Pb(NO3)2= 13 g NaNO3 x g NaNO3 170.0 g NaNO3 Only 8.5 grams of Na(NO3) can be formed!

  33. Calculating the Percent Yield • A batting average is actually a percent yield • On Base Hits/Official at bats

  34. Percent Yield • Percent yield = actual yield x 100 theoretical yield • Actual yield: the yield you ACTUALLY received. This is found experimentally. • Theoretical yield: the yield you SHOULD HAVE received. This is found mathematically.

  35. Calcium carbonate is decomposed by heating, as shown in the following equation: CaCO3 (s)  CaO (s) + CO2 (g) What is the percent yield if 13.1 grams CaO is produced from 24.8 g CaCO3 ? • First, find the theoretical yield of CaO in grams? • 24.8 g CaCO3=100.1 g CaCO3 x g CaO 56.1 g CaO = 13.9 g CaO

  36. CaCO3 (s)  CaO (s) + CO2 (g) • What is the percent yield if 13.1 grams CaO is produced? • Actual yield: 13.1 grams • Theoretical yield: 13.9 g CaO • Percent yield = actual yield x 100 theoretical yield

  37. % yield • % yield:13.1 g CaO x 100 13.9 g CaO = 94.2 %

  38. An experiment is performed where 31.5 g of CO reacts with 19.2 g of O2 to produce 48.2 g of CO2. What is the percent yield? • Balanced equation • Limiting reagent • Theoretical yield • Percent yield • 2 CO + O2 --> 2 CO2

  39. Limiting Reagent • 31.5 g CO =56.0 g CO= 18.0 g O2 x g O2 32.0 g O2 • 18.0 g of O2 is required to react with 31.5 g of CO • We have 19.2 g of O2, so the CO will run out first • CO is the limiting reagent

  40. Theoretical Yield • Use the limiting reagent • 31.5 g CO = 56.0 g CO = x g CO2 88.0 g CO2 • 49.5 g CO2

  41. Percent Yield • Percent yield = actual yield x 100 theoretical yield • 48.2 x 100 49.5 • 97.4 %

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