Algebraic Solving Method

1
Algebraic Solving Method

• Balancing Chemical Equations

2
Balance This Equation
Pb(N3)2 Cr(MnO4)2 ? Cr2O3 MnO2 Pb3O4 NO
3
Give Each Compound a Variable

• Pb(N3)2 Cr(MnO4)2 ? Cr2O3 MnO2 Pb3O4 NO
• Let Pb(N3)2 A
• Let Cr(MnO4)2 B
• Let Cr2O3 C
• Let MnO2 D
• Let Pb3O4 E
• Let NO F

4
Solve For Each of the Variables

• Pb(N3)2Cr(MnO4)2?Cr2O3MnO2Pb3O4NO
• Let A1
• Pb A3E
• A is the amount of Pb on the left side of the
  equation
• 3E is the amount on the left side

5
Solve For Each of the Variables
Pb(N3)2Cr(MnO4)2?Cr2O3MnO2Pb3O4NO

• Substitute in A
• Pb 13E ? 1/3E
• N 6x1F ? 6F
• Cr B2C or B/2C
• Mn 2BD or BD/2
• O 8B3C2D4EF
• 8B3(B/2)3(2B)4(1/3)6

• A1
• Pb A3E
• N 6AF
• Cr B2C
• Mn 2BD
• O 8B3C2D4EF

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Solve for Oxygen

• O 8B3B/24B4/36
• Find a lowest common multiple of 2 and 3
• Multiply each side by 6
• 6x 8B(3B/2)(4B4/3)6
• 48B9B24B836
• 48B33B44
• 15B44

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Simplify Your Fractions

• A1
• B44/15
• C(B/2)x(44/15)
• C44/30
• D2B
• D(2/1)x(44/15)
• D88/15
• E1/3
• F6

• Find a GCD and multiply each.
• A15
• B44
• C22
• D88
• E5
• F90

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Now You Have A Finished Equation

• Substitute each value into the equation.
• A15 D88
• B44 E5
• C22 F90
• 15Pb(N3)2 44 Cr(MnO4)2 ? 22Cr2O3 88MnO2
  5Pb3O4 90 NO

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