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5 3 Lecture 5 3 Lecture Presentation Transcript

  • 5.3 Balancing Chemical Equations Pages 161 - 165
  • Goal 1 Describe and balance chemical equations
    • Word equation
    Methane + Oxygen  Carbon dioxide + Water Reactants Products
    • Chemical Equations
      • Balanced to show conservation of mass
      • #Atoms per element go in = # Atoms per element that come out
    Important !!
    • Chemical equations
      • Use chemical formulas and coefficients
    CH 4 + 2 O 2  CO 2 + 2 H 2 O
    • Coefficients used to show
      • # of molecules/compounds
      • # of moles
      • Multiply subscripts by coefficients to determine the # of atoms
    CH 4 + 2 O 2  CO 2 + 2 H 2 O
    • C =
    • H =
    • O =
    • C =
    • H =
    • O =
    CH 4 + 2 O 2  CO 2 + 2 H 2 O # of atoms on left MUST = # of atoms on right Is this equation balanced?
  • CH 4 + 2 O 2  CO 2 + 2 H 2 O Yes! It’s balanced
    • H =
    • O =
    • H =
    • O =
    H 2 + O 2  H 2 O # of atoms on left MUST = # of atoms on right Is this equation balanced?
    • Uh-Oh!
    • H =
    • O =
    • H =
    • O =
    H 2 + O 2  H 2 O # of atoms on left MUST = # of atoms on right How can we balance it?
    • Rules for balancing chemical equations
      • 1. Only change coefficients
      • 2. Never never never ever change a subscript
      • 3. Multiply subscripts by coefficients
      • 4. Compounds FIRST
      • 5. Elements LAST
    • H =
    • O =
    • H =
    • O =
    __ H 2 + __ O 2  __ H 2 O # of atoms on left MUST = # of atoms on right Try it…
    • H = 4 atoms
    • O = 2 atoms
    • H = 4 atoms
    • O = 2 atoms
    _ 2 _ H 2 + __ O 2  _ 2 _ H 2 O # of atoms on left MUST = # of atoms on right Try it…
    • Mg =
    • O =
    • Mg =
    • O =
    __ Mg + __ O 2  __ MgO # of atoms on left MUST = # of atoms on right Let’s try it again…
    • Mg = 2 atoms
    • O = 2 atoms
    • Mg = 2 atoms
    • O = 2 atoms
    _2_ Mg + __ O 2  _2_ MgO # of atoms on left MUST = # of atoms on right How did you do?
    • C =
    • H =
    • C =
    • H =
    __ CH 4  __ C 3 H 8 + __ H 2 # of atoms on left MUST = # of atoms on right Let’s try it one more time…
    • C =
    • H =
    • C =
    • H =
    _ 3 _ CH 4  __ C 3 H 8 + _ 2 _ H 2 # of atoms on left MUST = # of atoms on right How did you do?