Fastest way to balance chemical equation
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Fastest way to balance chemical equation
While I know that guess and check is usually the typical method to balancing equation, are there tips on points I should look out for or things I should do first to make balancing complex equation faster?
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Re: Fastest way to balance chemical equation
start by balancing the atoms with the highest number of atoms in the reaction. Typically you leave balancing oxygen and/or hydrogen until last
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Re: Fastest way to balance chemical equation
Another tip is to balance any freestanding elements last because they're the easiest to manipulate.
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Re: Fastest way to balance chemical equation
When I am balancing an equation that contains O2, I usually leave that for last because it can simply be multiplied at the end with the rest of the equation balanced. Also, if it is a relatively long chemical equation I write each individual element on its own and its corresponding amount on both the reactants/products side and manipulate from there.
Ex) Al + HCl--> AlCl3 + H2
Al=1 Al=1
H=1 H=2
Cl=1 Cl=3
then you start balancing (in this case I would balance AlCl3 first since it has an odd number and we want to get rid of those to make it easier)
Balanced) (2)Al + (6)HCl--> (2)AlCl3 + (3)H2
Ex) Al + HCl--> AlCl3 + H2
Al=1 Al=1
H=1 H=2
Cl=1 Cl=3
then you start balancing (in this case I would balance AlCl3 first since it has an odd number and we want to get rid of those to make it easier)
Balanced) (2)Al + (6)HCl--> (2)AlCl3 + (3)H2
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Re: Fastest way to balance chemical equation
I'm not sure if this is the fastest way, but if you are stuck the easiest way to start is by making a system of equations. For example if you have the reaction HF + SiO2 --> SiF4 + H20 then you can set that up as
aHF + bSiO2 --> cSiF4 + dH20
For each element go through so
H: 1a= 2d
F: 1a= 4c
Si: 1b= 1c
O: 2b=1d
If you assume a=1 then
a=1
d=1/2
c=1/4
b=1/4
multiply these all by 4 so there are no fractions
a=4
d=2
c=1
b=1
those are the coefficients so
4HF + SiO2 --> SiF4 + 2H2
aHF + bSiO2 --> cSiF4 + dH20
For each element go through so
H: 1a= 2d
F: 1a= 4c
Si: 1b= 1c
O: 2b=1d
If you assume a=1 then
a=1
d=1/2
c=1/4
b=1/4
multiply these all by 4 so there are no fractions
a=4
d=2
c=1
b=1
those are the coefficients so
4HF + SiO2 --> SiF4 + 2H2
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Re: Fastest way to balance chemical equation
My T.A. showed us how to solve harder "balancing equations" problems using a system of equations. You basically replace all the stoichiometry coefficients with variables and write an equation for each element (based on how many moles there are, reactant=products). Then just substitute/eliminate to find the value of each variable!
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Fastest way to balance chemical equation
I similarly have trouble not with the simple chemical equations, but balancing the harder chemical equations. The test 1 we just took featured longer, more complicated chemical equations, and I didn't even know where to begin. It seemed like every element I balanced affected another and another, and it was a complicated endless cycle. Any tips for how to break down the most complicated ones, where there the same elements are featured multiple times?
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Re: Fastest way to balance chemical equation
for the equations that seem to never balance-meaning you feel like you just keep adding and adding coefficients, I think the fastest way would probably be to use a system of equations. to do this you would label the coefficients for each piece of the reaction as a through d, and come up with a system of equations. you then would say for example "let A = 1" and use substitution to find all the other coefficiets
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Re: Fastest way to balance chemical equation
A fast way to know whether to increase the stoichiometric coefficient of an equation is recognizing that an odd number of elements such as oxygen on one side does not match the even number of say O2 (oxygen as a diatomic molecule).
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Re: Fastest way to balance chemical equation
A fast way to know whether to increase the stoichiometric coefficient of an equation is recognizing that an odd number of elements such as oxygen on one side does not match the even number of say O2 (oxygen as a diatomic molecule).
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