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Which has more atoms: 17.0 grams of ammonia, NH 3 , or 72.9 grams of hydrogen chloride, HCl ?

Introduction to Stoichiometry Stoichiometry comes from the Greek words stoicheion , meaning “element,” and metron , meaning “measure.”. Which has more atoms: 17.0 grams of ammonia, NH 3 , or 72.9 grams of hydrogen chloride, HCl ?.

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Which has more atoms: 17.0 grams of ammonia, NH 3 , or 72.9 grams of hydrogen chloride, HCl ?

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  1. Introduction to StoichiometryStoichiometry comes from the Greek words stoicheion, meaning “element,” and metron, meaning “measure.”

  2. Which has more atoms: 17.0 grams of ammonia, NH3, or 72.9 grams of hydrogen chloride, HCl? 17.0 grams of ammonia is equivalent to one mole of ammonia which is equivalent to 4 atoms times 6.02 x 1023. With HCl72.9 grams is equal to two moles. Since you have 2 atoms in HCl that means that you have 2 x (2) 6.02 x 1023 atoms of HCl. They have the same number of atoms.

  3. Stoichiometry • Involves the mass relationships between reactants and products in a chemical reaction. • *** All stoichiometry calculations start with a balanced chemical equation.

  4. Stoichiometry Problems • You are given something. • Write that down. • Begin conversion factor. • Use molar mass to change from g-mol or mol-g • Use mole ratio to change from moles of one substance to moles of another

  5. Mole Ratio • A conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction. • This information is obtained directly from the balanced chemical equation • 2 Al2O3(l) 4 Al(s) + 3 O2(g) 2 mol Al2O3 4 mol Al 2 mol Al2O3 3 mol O2 4 mol Al 3 mol O2 or or or 4 mol Al . 2 mol Al2O3 3 mol O2. 2 mol Al2O3 3 mol O2 4 mol Al

  6. Molar Mass • The mass, in grams, of one mole of a substance • Recall from chapter 7, this information is obtained directly from the periodic table • Al2O3 = 102.0 g/mol Al = 27.0 g/mol O2 = 32.0 g/mol 102 g Al2O3 1 mol Al2O3 27 g Al 1 mol Al 32 g O2 1 mol O2 or or or 1 mol Al2O3 102 g Al2O3 1 mol Al 27 g Al 1 mol O2 32 g O2

  7. Reaction Stoichiometry Problems • 4 types of problems: • 1. mol mol • 2. mol mol mass • 3. mass mol mol • 4. mass mol mol mass

  8. Conversions of Quantities in Moles

  9. Sample Problem 9-1 • In a spacecraft, the carbon dioxide exhaled by astronauts can be removed by it’s reaction with lithium hydroxide, LiOH, according to the following chemical equation. • CO2(g) + 2 LiOH(s) Li2CO3(s) + H2O(l) • How many moles of lithium hydroxide are required to react with 20 mol of CO2, the average amount exhaled by a person each day? 20 mol ? mol 2 mol LiOH 1 mol CO2 20 mol CO2 x = 40 mol LiOH

  10. Additional Sample Problem #1 • The elements lithium and oxygen react explosively to form lithium oxide, Li2O. How many moles of lithium oxide will form if 2 mol of lithium react? 4 Li(s) + O2(g) 2 Li2O(s) 2 mol ? mol 2 mol Li2O 4 mol Li 2 mol Li x = 1 mol Li2O

  11. Additional Sample Problem #2 • Tetrachloroethylene, C2Cl4, is a dry-cleaning fluid made from acetylene, C2H2, and Cl2, according to this equation: C2H2(s) + 3Cl2(g) + Ca(OH)2(s) C2Cl4(s) + CaCl2(g) + 2H2O(s) How many moles of Cl2 are needed to react with 4 mol of C2H2? 4 mol ? mol 3 mol Cl2 1 mol C2H2 4 mol C2H2 x = 12 mol Cl2

  12. Conversions of Amounts in Moles to Mass

  13. Sample Problem 9-2 • In photosynthesis, plants use energy from the sun to produce glucose, C6H12O6, and oxygen from the reaction of carbon dioxide and water. What mass, in grams, of glucose is produced when 3.00 mol of water react with carbon dioxide? • 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) 3.00 mol ? g 1 mol C6H12O6 6 mol H2O 180.2 g C6H12O6 1 mol C6H12O6 90.1 g C6H12O6 3.00 mol H2O x x =

  14. Sample Problem 9-3 • What mass of carbon dioxide, in grams, is needed to react with 3.00 mol of H2O in the photosynthetic reaction described in Sample Problem 9-2? • 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) ? g 3.00 mol 6 mol CO2 6 mol H2O 44.0 g CO2 1 mol CO2 132 g CO2 3.00 mol H2O x x =

  15. Additional Sample Problem #1 • When sodium azide is activated in an automobile airbag, nitrogen gas and sodium are produced according to the equation: • 2NaN3(s) 2Na(s) + 3N2(g) • If 0.50 mol of NaN3 react, what mass in grams of nitrogen would result? 0.50 mol ? g 3 mol N2 2 mol NaN3 28.0 g N2 1 mol N2 21 g N2 0.500 mol NaN3 x x =

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