Balancing Chemical equations

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					Algebraic Solving Method
Balancing Chemical Equations

Balance This Equation

Pb(N3)2 + Cr(MnO4)2  Cr2O3+ MnO2 + Pb3O4 + NO

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

Solve For Each of the Variables
Pb(N3)2+Cr(MnO4)2Cr2O3+MnO2+Pb3O4+ NO
Let A=1

 Pb: A=3E * A is the amount of Pb on the left side of the equation *3E is the amount on the left side

 A=1 Pb: A=3E N: 6A=F Cr: B=2C Mn: 2B=D O: 8B=3C+2D+4E+F
 Substitute in A Pb: 1=3E  1/3=E N: 6x1=F  6=F Cr: B=2C or B/2=C Mn: 2B=D or B=D/2 O: 8B=3C+2D+4E+F 8B=3(B/2)+3(2B)+4(1/3)+ 6

Solve For Each of the Variables

Solve for Oxygen
O: 8B=3B/2+4B+4/3+6 Find a lowest common multiple of 2 and 3 Multiply each side by 6 [6x] 8B=(3B/2)+(4B+4/3)+6 48B=9B+24B+8+36 48B=33B+44 15B=44

Simplify Your Fractions
 A=1  B=44/15  C=(B/2)x(44/15) C=44/30  D=2B D=(2/1)x(44/15) D=88/15  E=1/3  F=6 Find a GCD and multiply each.  A=15  B=44  C=22  D=88  E=5  F=90

Now You Have A Finished Equation
Substitute each value into the equation. A=15 D=88 B=44 E=5 C=22 F=90

15Pb(N3)2 +44 Cr(MnO4)2  22Cr2O3+ 88MnO2 + 5Pb3O4 +90 NO

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