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Chapter 3: Chemical Reactions and the Earth’s Composition

Chapter 3: Chemical Reactions and the Earth’s Composition. Problems: 3.1-3.3, 3.5, 3.11-3.86, 3.95-3.115, 3.119-3.120, 3.122, 3.125-3.128, 3.132, 3.134, 3.136-3.138-3.141. 3.2 The Mole. Stoichiometry (STOY-key-OM-e-tree): quantitative study of reactants and products in a chemical reaction

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Chapter 3: Chemical Reactions and the Earth’s Composition

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  1. Chapter 3: Chemical Reactions and the Earth’s Composition Problems: 3.1-3.3, 3.5, 3.11-3.86, 3.95-3.115, 3.119-3.120, 3.122, 3.125-3.128, 3.132, 3.134, 3.136-3.138-3.141

  2. 3.2 The Mole Stoichiometry (STOY-key-OM-e-tree): quantitative study of reactants and products in a chemical reaction Interpreting a Chemical Equation H2 (g) + Cl2 (g)  2 HCl(g) 1 molecule 1 molecule 2 molecules It follows that any multiples of these coefficients will be in same ratio! 2 H2 (g) + O2 (g)  2 H2O(g) 1000 _____ molecule(s) _____ molecule(s) _____ molecule(s) N _____ molecule(s) _____ molecule(s) _____ molecule(s) Since N= Avogadro’s # = 6.0221023 molecules = 1 mole 2 H2 (g) + O2 (g)  2 H2O(g) _____ mole(s) _____ mole(s) _____ mole(s) Thus, the coefficients in a chemical equation give the mole ratios of reactants and products.

  3. Example Problem Consider the following: 2 C2H6 (g) + 7 O2 (g)  4 CO2 (g) + 6 H2O (g) 1. How many moles of O2 will react with 2.50 moles of C2H6? 2. How many moles of CO2 form when 3.50 moles of O2 completely react?

  4. Stoichiometric Calculations and the Carbon cycle Mass-Mass Stoichiometry Problems Example 1: Photosynthesis is the process of energy from sunlight being used to convert carbon dioxide into organic compounds, especially sugars like glucose, C6H12O6: 6 CO2(g) + 6 H2O(g) C6H12O6(aq) + 6 O2(g) What mass (in g) of glucose is produced via photosynthesis when 25.0 kg of carbon dioxide react with excess steam?

  5. 6 CO2(g) + 6 H2O(g) C6H12O6(aq) + 6 O2(g) What mass (in g) of glucose is produced via photosynthesis when 25.0 kg of carbon dioxide react with excess steam?

  6. Stoichiometric Calculations and the Carbon cycle Example 2: In biological systems, the reverse reaction, C6H12O6(aq) + 6 O2(g)  6 CO2(g) + 6 H2O(g) is called respiration and is the major source of energy for all livings things. a. Calculate the mass (in g) of carbon dioxide produced when 5.00 lb. of glucose reacts completely. (1 lb. = 453.6 g) b. How many pounds of carbon dioxide are produced in problem a. above?

  7. C6H12O6(aq) + 6 O2(g)  6 CO2(g) + 6 H2O(g) Calculate the mass (in g) of carbon dioxide produced when 5.00 lb. of glucose reacts completely. b. How many pounds of carbon dioxide are produced in problem a. above?

  8. 3.4 Combustion Reactions CxHy + O2(g) CO2(g) + H2O(g) CxHyOz+ O2(g)CO2(g) + H2O(g) Hydrocarbons (compounds with only C and H) and hydrocarbon derivatives (compounds with only C, H and O) burn in O2 to produce CO2 gas and steam, H2O(g).

  9. Combustion Reactions Example 1: Many home barbecues are fueled with propane gas (C3H8). Write the balanced equation for the combustion of propane, then calculate the mass (in kg) of carbon dioxide produced upon complete combustion of liquid propane from a 5.0 gal tank. (The density of liquid propane at 60°F is about 4.2 lbs. per gallon, and 1 lb. = 453.6 g)

  10. Combustion Reactions Example 2:Everclearis a brand of grain alcohol that can be as high as 190 proof (or 95% ethanol, C2H5OH, by volume). Calculate the mass of carbon dioxide produced upon complete combustion of the ethanol in a 750 mL bottle of Everclear. Write the balanced chemical equation for the combustion of ethanol. (The density of this Everclear is 0.80 g/mL.)

  11. 3.9 Limiting Reactants and Percent Yield • In practice, reactants will not always be present in the exact amounts necessary to be converted completely into products. • Some reactants (usually the more expensive) are only present in a limited supply, so these are almost always completely used up • “limiting reactant” (or limiting reagent)since it limits the amount of product made • Some reactants (usually the less expensive) are present in larger amounts and are never completely used up  “reactant(s) in excess”

  12. Guidelines for solving Limiting Reactant Problems • Calculate the mass or the # of moles of the 2nd reactant needed to completely reactwith the 1streactant. • If the moles needed is greater than the number of moles present for the 2nd reactant • That 2nd reactant will run out before the 1streactant. • The 2nd reactant = the limiting reactant, and the 1st reactant is in excess. • If the moles needed is less than the number of moles present for the reactant, • The 1st reactant = the limiting reactant, and the 2nd reactant is in excess. • Use the amount of the limiting reactant present to solve for the mass or # of moles of product that can be made.

  13. Limiting Reactant Problems Consider the reaction to produce ammonia: N2(g) + 3 H2(g)  2 NH3(g) Example 1: a) If 50.0 g of N2 react with 10.0 g of H2, what mass of ammonia is produced? b) The limiting reactant is _______ and the excess reactant is _________. c) What mass of the reactant in excess remains after the reaction?

  14. Calculating Percent Yield Percent yield =  100% • Theoretical yield: Amount of product one should get based on the chemical equation and the amount of reactants present • One generally calculates this in grams from info given • Actual yield: Amount of product one actually obtains • Generally smaller than the theoretical yield because of impurities and other adverse conditions in the lab • This is generally determined experimentally in the lab or given for a problem in lecture.

  15. Calculating Percent Yield Example 1:N2(g) + 3 H2(g)  2 NH3(g) a. For the reaction of 50.0 g of N2 with 10.0 g of H2, the theoretical yield of ammonia was determined to be what? theoretical yield = If 49.6 g of ammonia were actually produced, calculate the percent yield for the reaction. percent yield =

  16. Calculating Percent Yield Example 2: Consider the following reaction: 2 KClO3(s) 2 KCl(s) + 3 O2(g) What is the percent yield if 50.0 g of KClO3 decomposes to produce 16.4 g of oxygen gas?

  17. Calculating Percent Yield Example 3: Consider the following reaction: 3 Na2CrO4 (aq) + 2 AlCl3 (aq)Al2(CrO4)3 (s) + 6 NaCl(aq) a. What mass of precipitate is produced when 50.0 g of sodium chromate react with 50.0 g of aluminum chloride? Which is the limiting reactant and which is the reactant in excess?

  18. Calculating Percent Yield Example 3: Consider the following reaction: 3 Na2CrO4(aq) + 2 AlCl3(aq) Al2(CrO4)3(s) + 6 NaCl(aq). b. What mass of the reactant in excess remains after the reaction? c. What is the percent yield if 4.32 g of precipitate is actually produced?

  19. Calculating Percent Yield Example 4: Calculate the mass of methane (CH4) that must react to produce 10.0 kg of carbon dioxide if the percent yield for the reaction is 88.8%.

  20. Calculating Percent Yield Example 5: Consider the thermal decomposition of N2O5: 2 N2O5(g) 4 NO2(g) + O2(g) If the percent yield for the reaction is 96.8%, and the density of oxygen gas is 1.31 g/L, calculate the mass of N2O5 required to produce 50.0 L of oxygen gas.

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