Algebraic Solving Method : Algebraic Solving Method Balancing Chemical Equations
Balance This Equation : Balance This Equation Pb(N3)2 + Cr(MnO4)2 Cr2O3+ MnO2 + Pb3O4 + NO
Slide 3 : 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 Give Each Compound a Variable
Solve For Each of the Variables : Solve For Each of the Variables Pb(N3)2+Cr(MnO4)2Cr2O3+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
Solve For Each of the Variables : 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)2Cr2O3+MnO2+Pb3O4+NO
Solve for Oxygen : 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 : 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 : 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