che106_lecture6

che106_lecture6 - Chemistry 106 Lectures 6 Topics:...

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Unformatted text preview: Chemistry 106 Lectures 6 Topics: Reac7ons and Stoichiometry Lectures 6+7 = Chapter 3.1 - 3.7 Stoichiometry: Quan7ta7ve Rela7ons in Chemical Reac7ons •  Stoichiometry is the calcula7on of the quan77es of reactants and products involved in a chemical reac7on. –  It is based on balanced chemical equa7ons and on the rela7onship between mass and moles. –  Such calcula7ons are fundamental to most quan7ta7ve work in chemistry. Chemical Reac7ons: Equa7ons •  A chemical equa4on is the symbolic representa7on of a chemical reac7on in terms of chemical formulas. –  For example, the reac7on of sodium and chlorine to produce sodium chloride is wriOen: 2Na + Cl2 → 2NaCl •  The reactants are star7ng substances in a chemical reac7on. The arrow means “yields”. •  The formulas on the right side of the arrow represent the products. Anatomy of a Chemical Equa7on Methane reacts with oxygen to produce the flame in a Bunsen burner. Anatomy of a Chemical Equa7on methane plus oxygen yields carbon dioxide and water CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Anatomy of a Chemical Equa7on CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Reactants appear on the leV side of the equa7on. Anatomy of a Chemical Equa7on CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Products appear on the right side of the equa7on. Anatomy of a Chemical Equa7on CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) The states of the reactants and products are wriOen in parentheses to the right of each compound. Chemical Reac7ons: Equa7ons 2Na(s ) + Cl 2 (g ) → 2NaCl(s ) Ø།  g = gas Ø།  l = liquid Ø།  s = solid Ø།  aq = aqueous (in solu7on with water) Anatomy of a Chemical Equa7on CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) •  Coefficients are inserted to balance the equa7on. •  Balance just means that the total number of atoms of each element on both sides of a chemical equa7on must match. Subscripts and Coefficients Give Different Informa7on •  Subscripts tell the number of atoms of each element in a molecule. Subscripts and Coefficients Give Different Informa7on •  Subscripts tell the number of atoms of each element in a molecule •  Coefficients tell the number of molecules. iClicker Ques7on How many oxygen atoms are present in MgSO4 • 7 H2O? A.  4 oxygen atoms This is known as a his case, B.  5 oxygen atoms hydrate. In tevery it means that C.  7 oxygen atoms MgSO formula unit also has 7 H O w D. 11 oxygen atoms molecules o ieakly aOached t t. E.  18 oxygen atoms 4 2 Chemical Reac7ons: Balancing Equa7ons •  The law of conserva4on of mass dictates that the total number of atoms of each element on both sides of a chemical equa7on must match. •  The equa7on is then said to be balanced. •  Consider the combus7on of methane to produce carbon dioxide and water. CH 4 + O 2 → CO 2 + H 2 O Chemical Reac7ons: Equa7ons •  For this equa7on to balance, two molecules of oxygen must be consumed for each molecule of methane, producing one molecule of CO and two 2 molecules of water. CH 4 + 2 O 2 → CO 2 + 2 H 2O Now the equation is “balanced”. Balanced atom count: reactants: products: 1 C, 4 O, 4 H 1 C, 4 O, 4 H Chemical Reac7ons: Equa7ons • Balance the following equa7ons (via trial and error). • Coefficients are always the smallest whole number possible. O 2 + 2 PCl 3 → 2 POCl 3 P4 + 6 N 2O → P4O6 + 6 N 2 Ca 3 (PO4 )2 + 4 H 3 PO 4 → 3 Ca(H 2 PO 4 )2 2 As 2S 3 + 9 O 2 → 2 As 2O 3 + 6 SO 2 Combina7on Reac7ons •  In this type of reac7on, two or more substances react to form one product. •  Examples: Ø། 2 Mg (s) + O2 (g) ⎯ྎ⎯ྎ→ 2 MgO (s) Ø། N2 (g) + 3 H2 (g) ⎯ྎ⎯ྎ→ 2 NH3 (g) Ø། C3H6 (g) + Br2 (l) ⎯ྎ⎯ྎ→ C3H6Br2 (l) Combus7on Reac7ons •  These are generally rapid reac7ons that produce a flame. •  Hydrocarbons react with oxygen in the air to yield CO2 and H2O. •  Examples: Ø། CH4 (g) + 2 O2 (g) ⎯ྎ⎯ྎ→ CO2 (g) + 2 H2O (g) Ø། C3H8 (g) + 5 O2 (g) ⎯ྎ⎯ྎ→ 3 CO2 (g) + 4 H2O (g) Combus7on Reac7ons •  This box represents the total CO2 and H2O molecules formed by complete combus7on of a hydrocarbon. •  What is the empirical formula of the hydrocarbon? 1.  The unknown molecule is a hydrocarbon, so we only need to worry about C and H atoms. 2.  Count the total number of C atoms 3.  Count the total number of H atoms C4H16 CH4 Balancing Combus7on Reac7ons What about balancing the combus7on reac7on of a compound containing C, H, and O? CH3OH (l) + O2 (g) → CO2 (g) + H2O (g) C is balanced, O is balanced, but H is not. CH3OH (l) + O2 (g) → CO2 (g) + 2H2O (g) H is balanced now, but O is unbalanced. 3 O atoms on leV, 4 O atoms on right. How can we add one more O atom to the reactant side? Balancing Combus7on Reac7ons We can use a trick to help balance this chemical reac7on. CH3OH (l) + O2 (g) → CO2 (g) + 2H2O (g) Frac7ons may be used as intermediate steps in balancing. 3 CH3OH (l) + O2 (g) → CO2 (g) + 2H2O (g) 2 ALL balanced now, but frac7ons cannot be in the final answer! Last step: Mul7plying everything by 2 eliminates the frac7on. 2CH3OH (l) + 3O2 (g) → 2CO2 (g) + 4H2O (g) Decomposi7on Reac7ons •  In a decomposi7on, one substance breaks down into two or more substances. •  Examples: Ø། CaCO3 (s) ⎯ྎ⎯ྎ→ CaO (s) + CO2 (g) Ø། 2 KClO3 (s) ⎯ྎ⎯ྎ→ 2 KCl (s) + O2 (g) Ø། 2 NaN3 (s) ⎯ྎ⎯ྎ→ 2 Na (s) + 3 N2 (g) Decomposi7on Reac7ons Sodium azide is used for the infla7on of safety air bags in automobiles. 2 NaN3 (s) → 2 Na (s) + 3 N2 (g) ~100 grams (~0.2 lbs) of NaN3 yields ~50 liters of N2 gas in less than 50 ms (0.05 seconds)! Next Lecture •  Topics: Mass and Moles •  Text Reading: chapter 3 con7nued ...
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