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Stoichiometry

Stoichiometry. L. What is it really?. Cooking Method of finding how many cookies you can make given these ingredients how many ingredients you had to make this many cookies. All goes back to balanced equations. First write the equation and balance it.

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Stoichiometry

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  1. Stoichiometry L

  2. What is it really? • Cooking • Method of finding • how many cookies you can make given these ingredients • how many ingredients you had to make this many cookies

  3. All goes back to balanced equations • First write the equation and balance it. • You are observing the combustion of methane, CH4. Your teacher asks you how many moles of water are expected to form in the complete combustion of 2.50 mol of methane. CH4 + O2 CO2 + H2O UNBALENCED CH4 + 2O2 CO2 + 2H2O BALENCED

  4. Now the math • Remember that the balanced equations shows the relationship of how many moles of each compound is used in this reaction • Like the ingredients on a recipe (3 eggs + ½ cup water = eggy water) • So the coefficient of one compound to the next is the mole ratio

  5. Now the math (….continued) You are observing the combustion of methane, CH4. Your teacher asks you how many moles of water are expected to form in the complete combustion of 2.50 mol of methane. CH4 + 2O2 CO2 + 2H2O So if we had 1 mole of methane we would produce 2 moles of water But since we have 2.50 mol CH4…

  6. The math You are observing the combustion of methane, CH4. Your teacher asks you how many moles of water are expected to form in the complete combustion of 2.50 mol of methane. CH4 + 2O2 CO2 + 2H2O Therefore we would expect to retrieve 5 mols of H2O from the combustion of 2.5 mol of methane Mole ratio 5 mol H2O

  7. mole to mole • The synthesis reaction between nitrogen monoxide and oxygen forms nitrogen dioxide. • How many moles of nitrogen monoxide would be needed to fully react with 3.6 moles of oxygen • How many moles of nitrogen dioxide would be produced from 3.6 moles of oxygen • How many moles of oxygen would be needed to produce 4.67 mol nitrogen dioxide? • How many moles of nitrogen monoxide would be needed to produce 4.67 mol nitrogen dioxide?

  8. Building off that… • If you can do mol to mol, the rest should be child’s play • Mass to mol • Convert mass to mol, then mol to mol. • The end • Mass to mass • Convert mass of given to mol, convert mols of given to moles of wanted, convert mol to mass • The end • Particles to anything • Use Avagotto’s number (6.022 x 1023particles/mol)

  9. Mass to mol • In a reaction between the elements aluminum and chlorine, aluminum chloride is produced. How many grams of AlCl3 will be produced if 2.50 moles of Al react? Al + Cl2 AlCl3 Unbalanced 2Al + 3Cl2 2AlCl3 Balanced

  10. Mol to mass continued 2Al + 3Cl2 2AlCl3 How many grams of AlCl3 will be produced if 2.50 moles of Al react? Therefore 333.33 g of AlCl3 will be produced by the reaction of 2.50 mol of Al 333.33 g AlCl3 2.50 mol AlCl3 1 mol AlCl3

  11. Practice The ammonia (NH3) used to make fertilizers for lawns and gardens is made by reacting nitrogen and hydrogen. • Determine the mass in grams of NH3 formed from 1.34 moles of nitrogen. • What is the mass in grams of hydrogen required to react with 1.34 moles of nitrogen? • How many moles of nitrogen are required to produce 11.7 grams of NH3?

  12. Mass to mass • When nitrogen and hydrogen react, they form ammonia gas, which has the formula NH3. If 56.0 g of nitrogen are used up in the reaction, how many grams of ammonia will be produced? N2+ H2 NH3 Unbalanced N2+ 3H2 2NH3 Balanced

  13. When nitrogen and hydrogen react, they form ammonia gas, which has the formula NH3. If 56.0 g of nitrogen are used up in the reaction, how many grams of ammonia will be produced? N2 + 3H2 2NH3 • Convert grams to mol • Convert mol to gram Therefore, 68.11 g of NH3 are produced from 56 g of N2 3.9971 mol NH3 68.11 g NH3

  14. Particles to Particles • Hydrogen and Oxygen react to produce water. How many particles of hydrogen would be needed to react fully with 1.01 x 1012 particles of oxygen? H2+ O2 H2O Unbalanced 2H2 + O2  2H2O Balanced

  15. Hydrogen and Oxygen react to produce water. How many particles of hydrogen would be needed to react fully with 1.01 x 1012 particles of oxygen? 2H2 + O2  2H2O • Convert particles to mol • Convert mol to mol • Convert mol to particles Therefore we expect to need 2.14 x 1012 particles of H2 to react fully with 1.01 x 1012 particles of O2 2.14 x 1012 particles of H2 1.6772 x 10-12 mol O2

  16. Practice • Packet!

  17. Percent yield • How much are you actually getting out of your experiment? % Yield = Actual yield/theoretical yield x 100

  18. What is the total mass of H2O produced when 384g Cu is completely consumed in the following reaction?Cu + 4HNO3 Cu(NO3)2 + 2H2O + 2NO2 • Convert mass to mol • Convert mol to mol • Convert mol to mass Therefore we expect to produce 216g of H2O. What is the percent yield if this procedure was carried out in the lab and only produced 200g of H2O 216g H2O

  19. We expected to produce 216g of H2O.What is the percent yield if this procedure was carried out in the lab and only produced 200g of H2O • What was the actual yield? • 200g • What was the theoretical yield? • 216g • Solve • Therefore, our % yield is 92.6% 200g Actual yield 92.6% = % yield = Theoretical yield 216g

  20. Practice • Ethanol (C2H5OH) is produced from the fermentation of sucrose in the presence of enzymes: C12H22O11(aq) + H2O(g) C2H4OH(l) + CO2(g) • determine the theoretical and percent yields of ethanol if 684g sucrose undergoes fermentation and 349g ethanol is obtained T.Y. = 369g % yield = 94.6% • Lead (II) oxide is obtained by roasting galena, lead (II) sulfide, in air: PbS(s) + O2(g) PbO(s) + SO2(g) • Determine the theoretical and percent yields of PbO if 200.0g PbS is heated and 170.0g PbO is obtained T.Y. = 186.6g PbO % Yield = 91.10%

  21. Limiting reagent • The reactant that is completely consumed during a rxn • Reactants that remain are excess reactants Reagent

  22. To find the limiting reagent • Compare the actual mol:mol ratio to the balanced formula’s mol:mol ratio • Calculations must be based on given amounts of limiting reactants

  23. Fe + 2NiO(OH) + 2H2O  Fe(OH)2 + 2Ni(OH)2 • Determine the number of moles of Fe(OH)2 produced if 5.00 mol Fe and 8.00 mol of NiO(OH) react. What is the limiting reactant? What reactant is in excess and by how many moles? • 1mol Fe::2mol NiO(OH). Therefore,5mol Fe would need 10mol NiO(OH) • Since we only have 8.00mol NiO(OH), this is our limiting reagent.

  24. Fe + 2NiO(OH) + 2H2O  Fe(OH)2 + 2Ni(OH)2Determine the number of moles of Fe(OH)2 produced if 5.00 mol Fe and 8.00 mol of NiO(OH) react. What is the limiting reactant? What reactant is in excess and by how many moles? • Since NiO(OH) is the limiting reagent, find how many moles of Fe are needed to completely react with the amount of it we do have • 4 Mols of Fe are needed • Therefore 1 mol of Fe is in excess 4mol Fe 5.00mol Fe(provided) - 4mol Fe(needed) = 1mol Fe(excess)

  25. It gets complicated • If you are given anything other then mol, you need to first convert to mol for BOTH reagents before you can determine the limiting factor! Limiting Reagents

  26. Zn + 2MnO2 +H2O  Zn(OH)2 + Mn2O3 • Determine the limiting reactant if 25.0g Zn and 30.0g MnO2 are used, Determine the mass of Zn(OH)2 produced. • 0.77g moles of MnO2 required to react with 25.0g of Zn • 0.17 moles of Zn required to react with 30.0g of Mn(OH)2 • Mn(OH)2 is the limiting reagent • excess of 0.21mol Zn (0.38-0.17 = 0.21) • Use 30.0g to find the mass of Zn(OH)2 • 17.07g Zn(OH)2

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