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Stoichiometry

Stoichiometry. TJ Bautista, Sean Higgins, Joanna Lee Period 6. Definitions. Limiting Reactant/Limiting Reagent: the reactant that limits the amount of the other reactants that can combine and the amount of product that can form in the chemical reaction.

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Stoichiometry

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  1. Stoichiometry TJ Bautista, Sean Higgins, Joanna Lee Period 6

  2. Definitions Limiting Reactant/Limiting Reagent: the reactant that limits the amount of the other reactants that can combine and the amount of product that can form in the chemical reaction. Excess reactant: The substance that is not used up completely in a reaction

  3. How to Solve for Limiting Reagent 1. Write the balanced chemical equation 2. Determine the moles of each reactant 3. Determine how many moles of product EACH reactant would make using a mole ratio 4. The reactant that yields less products is the limiting reagent

  4. How does the limiting reactant affects the amount of products formed? The limiting reactant is the reactant that produces the smallest number of moles after all the calculations are performed

  5. 1) If 10.0 grams of NaOH react with 20.0 grams of H2SO4 to produce Na2SO4 , which reactant is limiting? NaOH + H2SO4 --> Na2SO4 + H2O

  6. 10.0g NaOH x 1 mole NaOH x 1 mole Na2SO4 = 0.125 mole Na2SO4 40g NaOH 2 mole NaOH 20.0g H2SO4 x 1 mole H2SO4 x 1 mole Na2SO4 = 0.204 mole Na2SO4 98.09g H2SO4 1 mole H2SO4 NaOH is the limiting reactant 1) If 10.0 grams of NaOH react with 20.0 grams of H2SO4 to produce Na2SO4 , which reactant is limiting? NaOH + H2SO4 --> Na2SO4 + H2O

  7. 2) If 15.0 grams of copper metal react with a solution containing 10.0 grams of AgNO3 , which reactant is limiting? Cu + AgNO --> Cu(NO ) + Ag 3 3 2

  8. 15.0g Cu x 1 mole Cu x 2 mole Ag = 0.472 mole Ag 63.54g Cu 1 mole Cu 10.0g AgNO3 x 1 mole AgNO3 x 2 mole Ag = 0.0589 mole Ag 169.91g AgNO3 2 mole AgNO3 AgNO3 is the limiting reactant 2) If 15.0 grams of copper metal react with a solution containing 10.0 grams of AgNO3 , which reactant is limiting? Cu + 2AgNO --> Cu(NO ) + 2Ag 3 3 2

  9. 3) A 2.00 grams sample of ammonia is mixed with 4.00 of oxygen. Which is the limiting reactant and how much excess reactant remains after the reaction has stopped? NH3(g) + O2(g) --> NO(g) + H2O(g)

  10. 3) A 2.00 grams sample of ammonia is mixed with 4.00 of oxygen. Which is the limiting reactant and how much excess reactant remains after the reaction has stopped? 4NH3(g) + 5O2(g) --> 4NO(g) + 6H2O(g)

  11. 2.00g NH3 x 1 mole NH3 x 4 mole NO x 30.0g NO = 3.53 g NO 17.0g NH3 4 mole NH3 1 mole NO 4.00g O2 x 1 mole O2 x 4 mole NO x 30.0g NO = 3.00 g NO 32.0g O2 5 mole O2 1 mole NO **O2 is the limiting reactant** 4.00g O2 x 1 mole O2 x 4 mole NH3 x 17.0g NH3 = 1.70 g NH3 32.0g O2 5 mole O2 1 mole NH3 2.00g NH3 (original sample) - 1.70g (reacted) = 0.30g NH3 (remaining) 4NH3(g) + 5O2(g) --> 4NO(g) + 6H2O(g) amount of ammonia that reacted

  12. 4) You have 22 grams of sodium nitrate and 16 g of sulfuric acid. Which reactant is in excess? How many grams is that reactant in excess? NaNO3(s) + H2SO4(l) --> Na2SO4(s) + HNO3(g)

  13. 4) You have 22 grams of sodium nitrate and 16 g of sulfuric acid. Which reactant is in excess? How many grams is that reactant in excess? 2NaNO3(s) + H2SO4(l) --> Na2SO4(s) + 2HNO3(g)

  14. 22g NaNO3 x 1 mole NaNO3 x 1 mole H2SO4 x 98g H2SO4 84.9g NaNO3 2 mole NaNO3 1 mole H2SO4 = 12.6g H2SO4 **NaNO3 is the limiting reactant** 16g Sulfuric Acid - 12.6g Sulfuric Acid = 3.4g Sulfuric Acid (remaining) 2NaNO3(s) + H2SO4(l) --> Na2SO4(s) + 2HNO3(g)

  15. Definitions Theoretical Yield: is the maximum amount of product that can be produced from a given amount of reactant Actual Yield: the measured amount of a product obtained from a reaction Percentage Yield: is the ratio of the actual yield to the theoretical yield times 100

  16. Percentage Yield 100 percentage = actual yield yield theoretical yield X

  17. 5) When 40.0 grams C6H6 react with an excess of Cl2 , the actual yield of C6H5Cl is 50g. What is the percentage yield of C6H5Cl? C6H6(l) + Cl2(g) --> C6H5Cl(l) + HCl(g)

  18. 40.0g C6H6 x 1 mole C6H6 x 1 mole C6H5Cl x 112.56g C6H5Cl 78.12g C6H6 1 mole C6H6 1 mole C6H5Cl = 57.6g C6H5Cl percentage yield = 40.0g X 100 = 69.4% 57.6g 5) When 40.0 grams C6H6 react with an excess of Cl2 , the actual yield of C6H5Cl is 50g. What is the percentage yield of C6H5Cl? C6H6(l) + Cl2(g) --> C6H5Cl(l) + HCl(g) theoretical yield

  19. 6) If 80.0 grams CO reacts to produce 70.0 grams CH3OH, what is the percentage yield of CH3OH? CO(g) + 2H2(g) -------------> CH3OH(l) catalyst

  20. 80.0g CO x 1 mole CO x 1 mole CH3OH x 32.04g CH3OH 28.01g CO 1 mole CO 1 mole CH3OH = 91.5g CH3OH percentage yield = 80.0g X 100 = 87.4% 91.5g 6) If 80.0 grams CO reacts to produce 70.0 grams CH3OH, what is the percentage yield of CH3OH? CO(g) + 2H2(g) -------------> CH3OH(l) catalyst theoretical yield

  21. Definitions Empirical Formula: gives the simplest whole-number ratio of the atoms of the elements Molecular Formula: shows the types and numbers of atoms combined in a single molecular compound

  22. 7) Find the empirical formula of a compound containing: 19.32 % Ca, 34.30 % Cl, and 46.38 % O.

  23. 7) Find the empirical formula of a compound containing: 19.32 % Ca, 34.30 % Cl, and 46.38 % O. 19.32g Ca x 1 mole Ca = 0.48204 mole Ca = 1 40.08g Ca 0.48204 34.30g Cl x 1 mole Cl = 0.96756 mole Cl = 2 35.45g Cl 0.48204 46.38g O x 1 mole O = 2.89875 mole O = 6 16.00g O 0.48204 mole ratio 1:2:6 Empirical formula = CaCl2O6 --> Ca(ClO3)2

  24. 8) Find the empirical formula of a compound containing: 64.86 % C, 13.52 % H, and 21.62 % O.

  25. 8) Find the empirical formula of a compound containing: 64.86 % C, 13.52 % H, and 21.62 % O. 64.86g C x 1 mole C = 5.4005 mole C = 1 12.01g Ca 1.35125 13.52g H x 1 mole H = 13.38614 mole Cl = 2 1.01g H 1.35125 21.62g O x 1 mole O = 1.35125 mole O = 6 16.00g O 1.35125 mole ratio 1:2:6 Empirical formula = C4H10O

  26. 9) Determine the molecular formula for a compound whose empirical formula is C5H7 and whose molecular mass is 268.44 g/mole.

  27. 9) Determine the molecular formula for a compound whose empirical formula is C5H7 and whose molecular mass is 268.44 g/mole. formula mass C5H7 = 67.11 g/mole 268.44 g/mole = 4 67.11 g/mole 4 x (C5H7) = C20H28

  28. 10) Find the molecular formula for a compound that contains 42.56g of palladium and 0.80 of hydrogen. The molecular mass of the compound is 433.68 g/mole.

  29. *Formula mass PdH2 = 108.42 g/mole 10) Find the molecular formula for a compound that contains 42.56g of palladium and 0.80 of hydrogen. The molecular mass of the compound is 433.68 g/mole. 42.56g Pd x 1 mole Pd = 0.4 mole Pd = 1 106.4g Pd 0.4 PdH2 0.80g H x 1 mole H = 0.79 mole H = 2 1.01g H 0.4 433.68 g/mole = 4 --------------> 4 x (PdH2) = Pd4H8 108.42 g/mole

  30. That concludes this presentation of Chapter 9 Stoichiometry! (: http://oformi.net/uploads/gallery/main/31/12319.jpg

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