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Balancing Chemical equations
Course: CHEM 307, Fall 2011School: Boise State
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Solving Method Method Balancing Algebraic Chemical Equations Balance This Equation Balance 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 Pb(N Cr(MnO Pb NO Let Pb(N3)2 = A Let Let Cr(MnO4)2 = B Let Let Cr2O3= C Let Let MnO2= D Let Let Pb3O4= E Let Let NO= F Let Solve For Each of the Variables Solve Pb(N3)2+Cr(MnO4)2Cr2O3+MnO2+Pb3O4+N...
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Solving
Method
Method
Balancing Algebraic Chemical Equations
Balance This Equation
Balance
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
Pb(N
Cr(MnO
Pb
NO
Let Pb(N3)2 = A
Let
Let Cr(MnO4)2 = B
Let
Let Cr2O3= C
Let
Let MnO2= D
Let
Let Pb3O4= E
Let
Let NO= F
Let
Solve For Each of the Variables
Solve
Pb(N3)2+Cr(MnO4)2Cr2O3+MnO2+Pb3O4+N
O
Let A=1
Pb: A=3E
* A iis the amount of Pb on the left side of the
s
equation
*3E is the amount on the left side
Solve For Each of the
Variables
Variables
Pb(N3)2+Cr(MnO4)2Cr2O3+MnO2+Pb3O4+NO
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
Pb:
1/3=E
N: 6x1=F 6=F
N:
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 Oxygen
O: for 8B=3B/2+4B+4/3+6
O:
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
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
Substitute each value into the equation.
A=15
D=88
B=44
E=5
C=22
F=90
C=22
F=90
15Pb(N3)2 +44 Cr(MnO4)2 22Cr2O3+
+44
88MnO2 + 5Pb3O4 +90 NO
88MnO 5Pb
+90
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