Balancing Chemical equations (Balancing Chemical Equations)Page 1

Slide 1

Algebraic Solving Method

Balancing Chemical Equations

Slide 2

Balance This Equation

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

Slide 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

Slide 4

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

Slide 5

Solve For Each of the Variables

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

Pb(N3)2+Cr(MnO4)2Cr2O3+MnO2+Pb3O4+NO

Slide 6

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

Slide 7

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

Slide 8

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