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Understanding Stoichiometry: Reagents, Products, and Balanced Equations

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daw
30 diapositives · 465 Views · Oct 11, 2014

Dive into the world of stoichiometry, the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. This guide covers the essential concepts of composition stoichiometry, reaction stoichiometry, and the balancing of chemical equations, including charge conservation. Learn to solve mole-mole stoichiometry problems and explore other problem types like volume-volume and mass-mass calculations. Master the use of balanced equations as conversion factors to predict the amounts of substances involved in reactions.

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  1. Stoichiometry Predicting amounts of reagents needed or amounts of products made

  2. Stoichiometry • Composition stoichiometry • Mass relationships of elements in compounds • Reaction stoichiometry • Mass relationships between participants in a chemical reaction • Must begin all stoichiometry problems with a BALANCED chemical equation.

  3. Conservation of Charge • Balanced chemical equation must be balanced for both mass and charge. • Total charge on reactant side must equal total charge on product side. • So far, most of the equations you have seen are neutral on each side, but this is not always the case.

  4. Coefficients in Balanced Equations • Give relative number of particles in rxn. • Give relative number of moles in rxn. • Coefficients give relationship between moles of each component in equation. • You use these relationships in MOLE-MOLE problems to convert from moles of 1 substance to moles of another substance.

  5. Stoichiometry Problems • Come in several flavors. • Start with the simplest: mole-mole • Given: amount of one substance in moles • Unknown: amount of some other substance in moles

  6. Mole-mole stoichiometry problem • Stoichiometry problems are factor-label problems or conversion problems. • Use the balanced chemical equation to generate the conversion factors.

  7. 2 C2H6 + 7 O2 4 CO2 + 6 H2O • How much CO2 will be produced if 4 moles of C2H6 are consumed? Identify a stoichiometry problem: 1) You always get a balanced chemical equation. 2) The problem will ask how many moles or how many liters of one species given a certain amount of another species. 3) The equation has a  arrow.

  8. 2C2H6 + 7 O2 4CO2 + 6 H2O • How much CO2 will be produced if 4 moles of C2H6 are consumed? • Identify given: 4 moles of C2H6. • Identify conversion factor from equation: 4 moles CO2 2 moles C2H6

  9. Conversion 4 moles C2H6 4 moles CO2 = 2 moles C2H6 4 X 4 moles CO2 = 8 moles CO2 2

  10. 2C2H6 + 7 O2 4CO2 + 6 H2O • How much H2O will be produced in the combustion of 5 moles of C2H6? • Given: 5 mol C2H6 • Conversion factor: 6 mol H2O 2 mol C2H6

  11. Conversion 5 mol C2H6 6 mol H2O = 2 mol C2H6 5 X 6 mol H2O = 15 mol H2O 2

  12. 2C2H6 + 7 O2 4CO2 + 6 H2O • How much oxygen will react with 5 moles of C2H6? • Given: 5 moles C2H6 • Conversion factor: 7 mol O2 2 mol C2H6

  13. Conversion 5 moles C2H6 7 mol O2 = 2 mol C2H6 5 X 7 mol O2 = 17.5 mol O2 2

  14. Other flavors of problems • Other types of stoichiometry problems include: • Volume-Volume (for gas phase only) • Mole-mass or mass-mole • Mass-Mass

  15. Gas-Phase Equations • 1 mole of any gas occupies the same volume as any other gas. At STP, 22.4 L. • So coefficients in equations represent the ratio of the volumes of gases involved in the reaction. • For rxns where all reactants & products are gases,  volume-volume problems. • Unit doesn’t matter as long as it’s constant throughout.

  16. N2(g) + 3 H2(g)  2 NH3(g) • How much hydrogen gas will react with 15 liters of nitrogen gas? • Given: 15 L of N2(g) • Conversion factor: 3 L H2 1 L N2

  17. Conversion 3 L H2 15 L N2(g) = 1 L N2 15 X 3 L H2 = 45 L H2 1

  18. Other types of stoich problems • General strategy: • Convert given into moles • Perform stoichiometric calculation using mole ratios from balanced equation • Convert to desired unit

  19. A word about … Double Replacement Reactions

  20. Reactions in aqueous solution • Many reactions, esp. many double replacement reactions, occur in water. • What happens when substances dissolve in water? • Depends on if they are ionic or covalent.

  21. Dissolving • Covalent substance – sugar or C6H12O6 • C6H12O6(s)  C6H12O6(aq) • The sugar molecules are spread out among the water molecules.

  22. Dissolving • Ionic substance – table salt or NaCl • NaCl(s)  Na+(aq) + Cl-(aq) • The ions are spread out among the water molecules.

  23. Double Replacement Reactions • Often occur when you mix 2 solutions of ionic compounds. • 1 product may be water or • 1 product may be a gas or • 1 product may be a solid • We say these reactions “go to completion.”

  24. Reactions producing Solids • Precipitation: the opposite of dissolving! • What do you see in the following clips: • S1043.mov • S1045.mov • S1046.mov • S1050.mov • S1057.mov • S1058.mov and S1060.mov

  25. Reactions producing Solids 2 NaOH(aq) + CuCl2(aq)  2 NaCl(aq) + Cu(OH)2(s) Complete Chemical Equation

  26. Complete Ionic Equations 2Na+(aq) + 2OH-(aq) + Cu2+(aq) + 2Cl-(aq)  2Na+(aq) + 2Cl-(aq) + Cu(OH)2(s) Substances that are ions in solution are written as ions in solution.

  27. Net Ionic Equations 2Na+(aq) + 2OH-(aq) + Cu2+(aq) + 2Cl-(aq)  2Na+(aq) + 2Cl-(aq) + Cu(OH)2(s) Notice that some ions do not participate in the reaction. They are spectator ions. Cross out all the spectator ions & you get the net ionic equation. 2OH-(aq) + Cu2+(aq)  Cu(OH)2(s)

  28. Reactions that form Water HBr(aq) + NaOH(aq)  H2O(l) + NaBr(aq) H+(aq) + Br-(aq) + Na+(aq) + OH-(aq)  H2O(l) + Na+(aq) + Br-(aq) H+(aq) + OH-(aq)  H2O(l)

  29. Reactions that form Gases HCl(aq) + NaHCO3(aq)  H2CO3(aq) + NaCl(aq) H2CO3(aq)  H2O(l) + CO2(g) ______________________________________ HCl(aq) + NaHCO3(aq)  H2O(l) + CO2(g) + NaCl(aq)

  30. Reactions that form Gases HCl(aq) + NaHCO3(aq)  H2O(l) + CO2(g) + NaCl(aq) H+(aq) + Cl-(aq) + Na+(aq) + HCO3-(aq)  H2O(l) + CO2(g) + Na+(aq) + Cl-(aq) H+(aq) + HCO3-(aq)  H2O(l) + CO2(g)