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Stoichiometry

Stoichiometry. Chapter 11. Stoichiometry = the study of quantitative relationships between the amounts of reactants used and products formed by a chemical reaction Based on the Law of Conservation of Mass Mass is neither created nor destroyed in a chemical reaction.

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Stoichiometry

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  1. Stoichiometry Chapter 11

  2. Stoichiometry = the study of quantitative relationships between the amounts of reactants used and products formed by a chemical reaction • Based on the Law of Conservation of Mass • Mass is neither created nor destroyed in a chemical reaction

  3. Law of Conservation of Mass 4Fe + 3O2 2Fe2O3 Reactants 4 mol Fe x 55.85g Fe = 223.4g Fe 1 mol Fe 3 mol O2 x 32.00g O2 = 96.00g O2 1 mol O2 Total 319.4 g Products 2 mol Fe2O3 x 159.7g Fe2O3 = 319.4 g Fe2O3 1 mol Fe2O3 Total 319.4 g

  4. Balancing Equations Review • Change the coefficients to make the number of atoms of each element equal on both sides of the equation *NEVER change asubscript…doing so changes the identity of the substance • Write the coefficients in their lowest possible ratio • Check your work

  5. A few things you do NOT do… H2 + O2 ---> H2O • You cannot change a subscript H2 + O2 ---> H2O2 2. You cannot place a coefficient in the middle of a formula H2 + O2 ---> H22O • Make sure that your final set of coefficients are all whole numbers with no common factors other than one 4 H2 + 2 O2 ---> 4 H2O

  6. Balancing Equations Review Balance this equation NaOH(aq) + CaBr2(aq)  Ca(OH)2(s) + NaBr(aq) 2NaOH(aq) + CaBr2(aq)  Ca(OH)2(s) + 2NaBr(aq)

  7. Review: Mole Ratio or (Conversion Factors) Mole Ratio = ratio between the numbers of moles of ANY two substances in a balanced chemical equation

  8. 2Al + 3Br22AlBr3 What mole ratios can be written for this reaction? 2mol Al3mol Br2 3molBr2 2mol AlBr3 2molAl2molAlBr3 2molAlBr3 2molAl 3mol Br22molAlBr3 2mol Al 3molBr2

  9. Helpful Hints • ALWAYS make sure the chemical equation is balanced!!! • Moles are ALWAYS involved when solving stoichiometric problems • The mole ratio is needed to convert from one substance to another in a balanced chemical equation • If you are in doubt about how to proceed in solving a problem, GO TO MOLES FIRST!!

  10. Mole to Mole Conversions

  11. How many moles of hydrogen are produced when 0.0400 mole of potassium is used? 2K + 2H20  2KOH + H2 • Identify the known… K • Identify the unknown…H2 • To solve this problem, you need to know how the unknown moles of H2 are related to the know moles of K • The correct ratio should have the moles of unknown in the numerator and the moles of the known should be in the denominator 1mol H2 2mol K

  12. This mole ratio can be used to convert the known number of moles of K to a number of moles of H2 0.0400mol K x 1mol H2 = 0.0200mol H2 2mol K If you put 0.0400mol K into water, 0.0200mol H2 will be produced

  13. Try one… C3H8 + 5O2 3CO2 + 4H2O How many moles of CO2 are produced when 10.0 moles of propane are burned in excess oxygen in a gas grill?

  14. Mole to Mass Conversions

  15. A mole to mass conversion is done when you know the number of moles of a reactant/product and you want to calculate the mass of another reactant/product

  16. Example: Determine the mass of NaCl produced when 1.25 moles of chlorine gas reacts vigorously with sodium. 2Na + Cl2 2NaCl • Unknown…NaCl • Known…Cl2 • Mole ratio…. 2mol NaCl 1 mol Cl2 • Multiply the given by the mole ratio 1.25mol Cl2 x 2mol NaCl = 2.50mol NaCl 1 mol Cl2

  17. Example: continued… Determine the mass of NaCl produced when 1.25 moles of chlorine gas reacts vigorously with sodium. 2Na + Cl2 2NaCl • Multiply the moles of NaCl by the molar mass of NaCl 2.50mol NaCl x 58.45g NaCl = 1mol NaCl = 146g NaCl produced

  18. Try one… Sodium chloride is decomposed into the elements sodium and chlorine by means of electrical energy. How many grams of chlorine gas can be obtained from 2.50mol NaCl? 2NaCl 2Na + Cl2

  19. Mass to Mass Conversions

  20. Example: Determine the mass of H2O produced from the decomposition of 25.0g of solid ammonium nitrate. NH4NO3 N2O + 2H2O • Unknown…H2O • Known… NH4NO3 • Convert the grams of the given to moles 25.0g NH4NO3 x 1mol NH4NO3 80.06g NH4NO3 = 0.312mol NH4NO3

  21. Example: Determine the mass of H2O produced from the decomposition of 25.0g of solid ammonium nitrate. NH4NO3 N2O + 2H2O 4. Mole ratio… 2mol H2O 1mol NH4NO3 5. Multiply by the moles of NH4NO3 previously calculated 0.312mol NH4NO3 x 2mol H2O = 0.624mol H2O 1mol NH4NO3

  22. Example: Determine the mass of H2O produced from the decomposition of 25.0g of solid ammonium nitrate. NH4NO3 N2O + 2H2O • Calculate the mass of H20 using the molar mass conversion factor 0.624mol H2O x 18.02g H2O = 11.2g H2O 1mol H2O

  23. Courtesy of Chemistry Matter and Change Glencoe McGraw-Hill Chemistry for Dummies John T. Moore www.karentimberlake.com

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