Chapter #3 Lectures Part Ii

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Chapter #3 Lectures Part Ii

  1. 1. g of O = g of sample – (g of C + g of H) Combust 11.5 g ethanol Collect 22.0 g CO 2 and 13.5 g H 2 O 6.0 g C = 0.5 mol C 1.5 g H = 1.5 mol H 4.0 g O = 0.25 mol O Empirical formula C 0.5 H 1.5 O 0.25 Divide by smallest subscript (0.25) Empirical formula C 2 H 6 O g CO 2 mol CO 2 mol C g C g H 2 O mol H 2 O mol H g H
  2. 2. A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction
  3. 3. How to “Read” Chemical Equations 2 Mg + O 2 2 MgO
  4. 4. Balancing Chemical Equations <ul><li>Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. </li></ul>Ethane reacts with oxygen to form carbon dioxide and water <ul><li>Change the numbers in front of the formulas ( coefficients ) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. </li></ul>C 2 H 6 + O 2 CO 2 + H 2 O
  5. 5. Balancing Chemical Equations <ul><li>Start by balancing those elements that appear in only one reactant and one product. </li></ul>
  6. 6. Balancing Chemical Equations <ul><li>Balance those elements that appear in two or more reactants or products. </li></ul>
  7. 7. Balancing Chemical Equations <ul><li>Check to make sure that you have the same number of each type of atom on both sides of the equation. </li></ul>
  8. 8. <ul><li>Write balanced chemical equation </li></ul><ul><li>Convert quantities of known substances into moles </li></ul><ul><li>Use coefficients in balanced equation to calculate the number of moles of the sought quantity </li></ul><ul><li>Convert moles of sought quantity into desired units </li></ul>Mass Changes in Chemical Reactions
  9. 9. Stoichiometric Calculations <ul><li>From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant) </li></ul>
  10. 10. Methanol burns in air according to the equation If 209 g of methanol are used up in the combustion, what mass of water is produced? 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O
  11. 11. Limiting Reagents
  12. 12. Limiting Reactants <ul><li>The limiting reactant is the reactant present in the smallest stoichiometric amount </li></ul>
  13. 13. Theoretical Yield <ul><li>The theoretical yield is the amount of product that can be made </li></ul><ul><ul><li>In other words it’s the amount of product possible as calculated through the stoichiometry problem </li></ul></ul><ul><li>This is different from the actual yield, the amount one actually produces and measures </li></ul>
  14. 14. In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 Calculate the mass of Al 2 O 3 formed. 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe
  15. 15. Use limiting reagent (Al) to calculate amount of product that can be formed.
  16. 16. Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction.
  17. 17. Lecture Problem #1 <ul><li>A reaction mixture contains 23.5 grams of phosphorus trichloride, PCl 3 , and 14.56 grams of lead (II) fluoride, PbF 2 . What mass of PbCl 2 can be obtained from the following reaction? </li></ul><ul><li>3 PbF 2 + 2 PCl 3 -> 2 PF 3 + 3 PbCl 2 </li></ul>
  18. 18. Lecture Problem #1 Solution <ul><li>1 mole of PCl 3 = 137.32 grams </li></ul><ul><li>23.5 grams PCl 3 = 0.171 moles PCl 3 </li></ul><ul><li>1 mole PbF 2 = 245.2 grams </li></ul><ul><li>14.56 grams PbF 2 = 0.0594 moles </li></ul>
  19. 19. Lecture Problem #2 <ul><li>What is the percent yield if 108 milligrams of sulfur dioxide, SO 2 , is isolated from the combustion of 82.7 mg of carbon disulfide, CS 2 , according to the reaction: </li></ul><ul><li>CS 2 + 3 O 2 -> CO 2 + 2 SO 2 </li></ul>
  20. 20. Lecture Problem #2 Solution <ul><li>1 mole CS 2 = 76.15 grams </li></ul><ul><li>82.7 mg CS 2 = 1.09 x 10 -3 moles CS 2 </li></ul><ul><li>1 mole CS 2 yields 2 moles of SO 2 </li></ul><ul><li>1 Mole SO 2 = 64.07 grams </li></ul>
  21. 21. Lecture Problem #3 <ul><li>The percent yield of the following reaction is consistently 78%. </li></ul><ul><li>CH 4(g) + 4 S (g) -> CS 2(g) + 2 H 2 S g) </li></ul><ul><li>How many grams of sulfur would be required to obtain 56.0 grams of CS 2 ? </li></ul>

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