June 07, 2020, 01:40:20 AM
Forum Rules: Read This Before Posting

### Topic: Need help with the algebraic method of balancing equations  (Read 145 times)

0 Members and 1 Guest are viewing this topic.

#### Cocktopus

• Very New Member
• • Posts: 2
• Mole Snacks: +0/-0 ##### Need help with the algebraic method of balancing equations
« on: May 19, 2020, 02:08:00 PM »
For simple problems, using the algebraic method of solving chemical equations has always been relatively easy.
Find the equations for each element, set a=1, solve for another variable, use that variable to solve for another, and so on down the line.

But it's harder for me now, because the problems are getting more complex.

Like this one: CaCO3+CH3COOH---------->Ca(CH3COO)2+H2O+CO2

Now when I try to solve, I get the following equations:

Ca: a=c
C: a+2b=4c+e
O: 3a+2b=4c+d+2e
H: 4b=6c+2d

Now I set a=1, and therefore I find c (as per the first equation), but the rest of the coefficients are harder to solve for. Which coefficients should I solve for first, and why? And how do I go about doing this?

#### AWK

• Retired Staff
• Sr. Member
• • Posts: 7331
• Mole Snacks: +515/-86
• Gender:  ##### Re: Need help with the algebraic method of balancing equations
« Reply #1 on: May 19, 2020, 02:45:03 PM »
Substitute a and c for the remaining equations - you have 3 equations with three unknowns. Then it's easiest to eliminate d.
AWK

#### Cocktopus

• Very New Member
• • Posts: 2
• Mole Snacks: +0/-0 ##### Re: Need help with the algebraic method of balancing equations
« Reply #2 on: May 19, 2020, 03:41:03 PM »
Could you also do it with substitution?

#### AWK

• Retired Staff
• Sr. Member
• • Posts: 7331
• Mole Snacks: +515/-86
• Gender:  ##### Re: Need help with the algebraic method of balancing equations
« Reply #3 on: May 19, 2020, 04:03:59 PM »
Subtract half of the last equation from the penultimate one. You get the value of e.
AWK

#### Borek ##### Re: Need help with the algebraic method of balancing equations
« Reply #4 on: May 19, 2020, 04:08:43 PM »
Could you also do it with substitution?

As a=c you can substitute a for c (or c for a) in all other equations, and bang - you have just reduced number of remaining variables to solve for.

Honestly, this is more about math than chemistry. There are no special tricks - you have to solve many problems to get some experience, then it becomes easy. It is typically easier to start with unknowns present in the simplest equations.

Substituting an integer for one of the unknowns was always the final step for me, once I eliminated all but the last two variables. Then it is easier to avoid nasty fractions.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info