Balancing Equation
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 9
- Joined: Fri Apr 06, 2018 11:04 am
Balancing Equation
Does anyone have any tips for balancing chemical equations in which the coefficients just seem to keep getting higher and whenever one element is balanced, it becomes unbalanced as you try balancing the next element?
-
- Posts: 59
- Joined: Wed Nov 15, 2017 3:01 am
Re: Balancing Equation
Well, in general, it's important to remember the basic rules and start balancing the least abundant element in a chemical equation to avoid this problem. If you still encounter it, however, you should just use a fraction to balance it temporarily. For example, let's say you need 19 Fe, but that means you will have to change many stoichiometric coefficients to even higher numbers. Then, you should use 19/2 Fe and then multiply the entire equation by 2 to get rid of the fraction and obtain the correct stoichiometric coefficients for the rest of the compounds/elements. Hope this helped :)
-
- Posts: 30
- Joined: Fri Apr 06, 2018 11:02 am
-
- Posts: 30
- Joined: Fri Apr 06, 2018 11:04 am
- Been upvoted: 1 time
Re: Balancing Equation
If the balancing question is not easy enough to just guess and check then you can use a method where you assign a variable to each unknown coefficient and then make a system of equations for how many times each element appears and solve the system.
For example, in __a_KClO3+__b_C6H12O6+__c__KCl+__d___CO2+__e__H20
K: 1a+0b=1c+0d+0e==>a=c
Cl: 1a+0b=1c+0d+0e==>a=c
O: 3a+6b=2d+e
C: 6b=d
H: 6b=e
You then assume that one variable=1 (in this case, lets make a=1) because there are more variables than equations. Then solve the system of equations.
You get: a=1, b=1/4, c=1, d=6/4, e=6/4, but because you cannot have fractions you multiply by a number that would get rid of all fractions which in this case is 4.
So the final is, a=4, b=1, c=4, d=6, e=6
For example, in __a_KClO3+__b_C6H12O6+__c__KCl+__d___CO2+__e__H20
K: 1a+0b=1c+0d+0e==>a=c
Cl: 1a+0b=1c+0d+0e==>a=c
O: 3a+6b=2d+e
C: 6b=d
H: 6b=e
You then assume that one variable=1 (in this case, lets make a=1) because there are more variables than equations. Then solve the system of equations.
You get: a=1, b=1/4, c=1, d=6/4, e=6/4, but because you cannot have fractions you multiply by a number that would get rid of all fractions which in this case is 4.
So the final is, a=4, b=1, c=4, d=6, e=6
-
- Posts: 31
- Joined: Fri Apr 06, 2018 11:02 am
Re: Balancing Equation
Personally, I like to look for the element that is present the least on both sides as a starting point. From there I try to start with the elements that have smaller quantities present on both sides of the reaction. In addition, like the example Dr. Lavelle used in class I will utilize fractions and multiply them to make it easier in the end.
-
- Posts: 96
- Joined: Fri Apr 06, 2018 11:05 am
- Been upvoted: 1 time
-
- Posts: 142
- Joined: Wed Nov 15, 2017 3:03 am
-
- Posts: 62
- Joined: Fri Sep 28, 2018 12:19 am
- Been upvoted: 1 time
Re: Balancing Equation
Other than beginning with the least abundant element and using inspection, you can also try the proportions method. This is similar to the matrix method, where you put variables to represent each coefficient in the equation. Then, you solve it like a mathematical equation with multiple variables. Hope this helped!
Return to “Balancing Chemical Reactions”
Who is online
Users browsing this forum: No registered users and 5 guests