El Segundo High School students that need chemistry tutoring are in luck. Study Hut tutoring in El Segundo is going to get you going where you want to go. We are experts in every subject, with tutors who’s knowledge is and teaching style is tailored to each of the students who come in for subject help and test prep. Take a look at this breakdown from Andy, our UCLA bio/chem major who dominates science like nobodies business (but ours!):
Sometimes balancing chemical equations is easy….unfortunately it can also be
a nightmare! When all else fails there is one way to balance an equation that will
always work; with ALGEBRA! Who ever said learning math wasn’t useful?
To solve balance chemical equations using algebra start by assigning a variable
to each part of the reaction. For example if we had the unbalanced equation
NaHCO2 + C6H8O7 ——-> CO2 + H2O + Na3C6H5O7
First we assign a value of 1 as the coefficient of the first compound. We then give
letter designations to the other coefficients.
(1)NaHCO2 + (A)C6H8O7 ——-> (B)CO2 + (C)H2O + (D)Na3C6H5O7
Now we can establish relationships between the variables based on the different
atoms.
Sodium: 1=3D
Hydrogen: 1 + 8A = 2C + 5D
Carbon: 1 + 6A= B + 6D
Oxygen: 2 + 7A = 2B + C + 7D
Now we solve these equations starting with Sodium
1=3D ——–> D=1/3
We then plug D=1/3 into the Hydrogen and Oxygen equations and solve for C
and B respectively:
1 + 8A = 2C + 5D ———> C = 4A – 1/3
1 + 6A = B + 6D ———–> B = 6A -1
Finally if we put these into the Oxygen equation we can solve for A:
2 + 7A = 2B + C + 7D ———–> 2 + 7A = 2(6A-1) + (4A-1/3) + 7(1/3)
This give A = 2/9. Now we can go back to Hydrogen and Oxygen and solve for C
and D.
C = 4A – 1/3 —–> 4(2/9) – 1/3 = C
C = 5/9
B = 6A -1 ——–> 6(2/9) -1 = B
B= 3/9
This give us: (1)NaHCO2 + (2/9)C6H8O7 ——-> (3/9)CO2 + (5/9)H2O + (1/3)
Na3C6H5O7
To get whole numbers we multiply by 9 giving us the solution!
(9)NaHCO2 + (2)C6H8O7 ——-> (3)CO2 + (5)H2O + (3)Na3C6H5O7
This method may seem complex, but it is guaranteed to work for any chemical
equation!
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