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Warm-up 5/3/11

Warm-up 5/3/11. Using the following, write a balanced equation: “B” – Bicycle “S” – Seat “P” – Peddles “W” – Wheels “F” – Frame “H” - Handlebars. Objective. Stoichiometry – calculations and notes. Answer to warm-up. 1S + 2p + 2W + 1F + 1H  1 B.

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Warm-up 5/3/11

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  1. Warm-up 5/3/11 • Using the following, write a balanced equation: • “B” – Bicycle • “S” – Seat • “P” – Peddles • “W” – Wheels • “F” – Frame • “H” - Handlebars

  2. Objective • Stoichiometry – calculations and notes

  3. Answer to warm-up • 1S + 2p + 2W + 1F + 1H  1 B

  4. Copy the following on a separate piece of paper.

  5. Unit 9 Stoichiometry Mole-Mole, Mass-Mass, Mass-Volume, Volume-Volume, Limiting Reactant and Percent Yield

  6. Stoichiometry • The study of quantitative, or measurable relationships that exist in chemical formulas and chemical reactions. • One form of stoich involves converting among moles, particles, mass and volume of chemical formulas. (Unit 8) • The other form of stoich is concerned with chemical reactions and involves the relationships between reactants and products in a chemical reaction. (Unit 9)

  7. Mole-Mole Problems • The coefficients in a balanced chemical equation can be interpreted both as the number of particles and as the number of moles involved in the reaction. • Observe the chemical equation below. 2H2 + O2 2H2O • According to the equation: 2 moles of hydrogen and one mole of oxygen produce 2 moles of water.

  8. Mole-Mole Problems Suppose you wanted to know how many moles of hydrogen and oxygen are needed to produce 5.48 moles of water. • First determine the molar ratios indicated by the equation. 2H2 + O2 2H2O • H2 to H2O is 2:2 • O2 to H2O is 1:2

  9. Mole-Mole Problems • Second, set up equations to convert the number of moles of the reactant to the number of moles of each of the products using dimensional analysis. • 5.48 mol H2O 2 mol H2 = 5.48 mol H2 2 mol H2O • 5.48 mol H2O 1 mol O2 = 2.74 mol O2 2 mol H2O

  10. Mass of Unknown Mass of given Moles of Unknown Moles of given Mass-Mass Problems You will be given the mass of one substance and are asked to find the mass of another substance involved in the same chemical reaction.

  11. Mass-Mass Problems • Determine the mass of NaOH produced when 0.25g of Na reacts with water according to the following equation: 2Na + 2H2O  2NaOH + H2 0.25g Na 1 mol Na 2 mol NaOH 39.99711g NaOH = 22.98977gNa 2 mol Na 1 mol NaOH 0.43g NaOH

  12. Mass-Mass Problems • How many grams of aluminum chloride will be produced from 92.00g of Cl2? 2AlBr3 + 3Cl2 3Br2 + 2AlCl3 92.00g Cl2 1 mol Cl2 2 mol AlCl3 133.341g AlCl3 = 70.906g Cl2 3 mol Cl2 1 mol AlCl3 115.3g AlCl3

  13. Volume of Gaseous Unknown Mass of given Moles of Unknown Moles of given Mass-Volume Problems • Used to calculate the volume of gas produced when the mass of another substance in the reaction is known.

  14. Mass-Volume Problems • How many liters of oxygen are necessary for the combustion of 340.00g of ethanol (C2H5OH) assuming that the reaction occurs at STP? C2H5OH + 3O2 2CO2 + 3H2O 340.00g C2H5OH 1 mol C2H5OH 3 mol O2 22.4 L O2 = 46.069g C2H5OH 1 mol C2H5OH 1 mol O2 495.95 L O2

  15. Mass-Volume Problems • How much hydrogen gas, at STP, if formed from the reaction of 3.50g of zinc with an excess amount of hydrochloric acid? Zn + 2HCl  ZnCl2 + H2 3.50 g Zn 1 mol Zn 1 mol H2 22.4 L H2 = 65.39g Zn 1 mol Zn 1 mol H2 1.20 L H2

  16. Volume-Volume Problems • The coefficients in a chemical equation represent the ratio of the volumes of gases involved in the reaction. • Very similar to Mole-Mole problems.

  17. Volume-Volume Problems What volumes of hydrogen and nitrogen gases are necessary to produce 16.0 L of ammonia gas? N2 + 3H2 2NH3 16.0 L NH3 1 mol NH3 1 mol N2 22.4 L N2 = 8.00 L N2 22.4 L NH3 2 mol NH3 1 mol N2 16.0 L NH3 1 mol NH3 3 mol H2 22.4 L H2= 24.0 L H2 22.4 L NH3 2 mol NH3 1 mol H2

  18. Limiting Reactants • The amount of product that can be produced from a chemical reaction depends on the amounts of reactants available. • Stoichiometric proportion – when the quantities of reactants are available in the exact ratio described by the balanced equation.

  19. Limiting Reactants • There is usually more of one reactant than can be used… this reaction is said to be in nonstoichiometric proportions. • The reactant that gets used up first and therefore limits the amount of product formed is called the limiting reactant.

  20. Limiting Reactants • To solve a limiting reactant problem, you must complete two separate mass-mass problems. First calculate the mass of product formed using the given mass of one reactant. Then calculate the mass of product formed using the given mass of the other reactant. • The limiting reactant is the one that produces the smaller amount of product.

  21. Limiting Reactant Problems • Identify the limiting reactant when 10.0 g H2O reacts with 3.5g Na to produce NaOH and H2. 2Na + 2H2O  2NaOH + H2 10.0 g H2O 1 mol H2O 2 mol NaOH 39.9971 g NaOH = 18.0153g H2O 2 mol H2O 1 mol NaOH 22.2 g NaOH

  22. Limiting Reactant Problems • Identify the limiting reactant when 10.0 g H2O reacts with 3.5g Na to produce NaOH and H2. 2Na + 2H2O  2NaOH + H2 3.5g Na 1 mol Na 2 mol NaOH 39.9971 g NaOH = 22.98977g Na 2 mol Na 1 mol NaOH 6.1 g NaOH

  23. Limiting Reactant Problems • NaCl reacts with Pb(NO3)2 to produce PbCl2 and NaNO3. What is the limiting reactant when 45.3g of NaCl reacts with 60.8g of Pb(NO3)2? Pb(NO3)2 + 2NaCl  PbCl2 + 2NaNO3 60.8g Pb(NO3)2 1 mol Pb(NO3)2 2 mol PbCl2 278.1054g PbCl2 = 331.2098g Pb(NO3)2 1mol Pb(NO3)2 1 mol PbCl2 102g PbCl2

  24. Limiting Reactant Problems • NaCl reacts with Pb(NO3)2 to produce PbCl2 and NaNO3. What is the limiting reactant when 45.3g of NaCl reacts with 60.8g of Pb(NO3)2? Pb(NO3)2 + 2NaCl  PbCl2 + 2NaNO3 45.3g NaCl 1 mol NaCl 1 mol PbCl2 278.1054g PbCl2 = 58.44247g NaCl 2 mol NaCl 1 mol PbCl2 108g PbCl2

  25. Limiting Reactant Problems • Identify the limiting rectant when 12.5 L H2S at STP is bubbled through a solution containing 24.0g KOH to form K2S and H2O. H2S + 2KOH  K2S + 2H2O 12.5 L H2S 1 mol H2S 2 mol H2O 18.01528g H2O =20.1g H2O 22.4L H2S 1 mol H2S 1 mol H2O 24.0g KOH 1 mol KOH 2mol H2O 18.01528g H2O = 7.71gH2O 56.10564g KOH 2mol KOH 1 mole H2O

  26. Percent Yield • The percent of the expected yield which was actually obtained experimentally. Percent Yield = actual yield x 100 expected yield Actual Yield- the amount of a product that is really obtained from a chemical reaction. Expected Yield- the amount of product that should be produced based on calculations.

  27. Percent Yield • Determine the percent yield for the reaction between 2.80g Al(NO3)3 and excess NaOH if 0.966g of Al(OH)3 is recovered. Al(NO3)3 + 3NaOH  Al(OH)3 + 3NaNO3 2.80gAl(NO3)3 1 mol Al(NO3)3 1mol Al(OH)3 78.00356gAl(OH)3 = 212.996gAl(NO3)3 1mol Al(NO3)3 1mol Al(OH)3 1.03g Al(OH)3

  28. Percent Yield Actual Yield = 0.966g Al(OH)3 Theoretical Yield = 1.03g Al(OH)3 % Yield = Actual Yield x 100 Theoretical Yield % Yield = 0.966g x 100 = 93.8% 1.03g

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