Balancing Chemical Equations Using Algebraic Solving Method
Algebraic Solving
Method
Balancing Chemical Equations
Balance This Equation
Pb(N3)2 + Cr(MnO4)2
Cr2O3+ MnO2 +
Pb3O4 + NO
Pb(N
3
)
2
+ Cr(MnO
4
)
2
Cr
2
O
3
+ MnO
2
+ Pb
3
O
4
+
NO
Let
Pb(N
3
)
2
= A
Let Cr(MnO
4
)
2
= B
Let
Cr
2
O
3
= C
Let MnO
2
= D
Let Pb
3
O
4
= E
Let NO= F
Give Each Compound a Variable
Give Each Compound a Variable
Solve For Each of the Variables
Pb(N
3
)
2
+Cr(MnO
4
)
2
Cr
2
O
3
+MnO
2
+Pb
3
O
4
+
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
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(N
3
)
2
+Cr(MnO
4
)
2
Cr
2
O
3
+MnO
2
+Pb
3
O
4
+NO
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(N
3
)
2
+44 Cr(MnO
4
)
2
22Cr
2
O
3
+
88MnO
2
+ 5Pb
3
O
4
+90 NO
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Learn how to balance chemical equations using the algebraic solving method step-by-step by assigning variables to compounds and solving for each variable. This comprehensive guide takes you through the entire process, from assigning variables to simplifying fractions and substituting values back into the equation. Master the art of balancing complex chemical equations effectively with this detailed tutorial.
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Presentation Transcript
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
Solve For Each of the Variables Pb(N3)2+Cr(MnO4)2 Cr2O3+MnO2+Pb3O4+NO 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 A=1 Pb: A=3E N: 6A=F Cr: B=2C Mn: 2B=D O: 8B=3C+2D+4E+F
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 B=44 C=22 D=88 E=5 F=90 15Pb(N3)2 +44 Cr(MnO4)2 22Cr2O3+ 88MnO2 + 5Pb3O4 +90 NO
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