[Eclipsonix]

I'm having the hardest trouble ever balancing equations for some reason >,<

I can't do this worth crap after 2 questions lol: http://funbasedlearning.com/chemistry/chemBalancer/ques9.htm

I'm curious what type of math do I have to know? I'm pretty rusty on my math. It's been a while, and I don't know what to learn in order to do this, and things like:


Thanks guys, again 

[opti384]

So is this the equation you're having trouble with?

_KMnO₄  _K₂O +_MnO + _O₂




This might give you some help http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients

[Eclipsonix]

I just really don't know what math I need to be learning in order to solve these chemical equations >,<

[skyjumper]

Algebra if anything. Practice, its more an acquired ability, i've found

[LPH]

Addition, subtraction, and multiplication are the basics for balancing an equation. Knowing common multipliers is also helpful. Finally, understanding how to get rid of a fraction will be helpful.

The key to balancing equations is being able to count atoms and making sure the number of atoms on both sides of the arrow are the same.

_KMnO₄    _K₂O +_MnO + _O₂

Consider the problem above and you'll see the three elements K, Mn, and O are involved.

Now, to get the total number of atoms of each element, you would multiply the coefficient (the number in front of the molecule) by the subscript. The _ in front of the molecule is a great way to indicate that these are blank spaces requiring the person to place numbers (coefficients) there.

We start with one K, 1 Mn, and 4 O on the left side of the arrow.
We have two K, 1 Mn, and 4 O on the right side of the arrow.

To balance, I like to use the ping-pong method and bounce back and forth until everything is balanced.

Because we have 2 K on the right, we'd need to balance this on the left. We do this by placing a coefficient of 2 in front of the molecule.

2 KMnO₄

Now, we have the correct number of K, but are incorrect with Mn (2) and O (.

Therefore we work on the right side and see that we can place a 2 in front of the molecule MnO and this balances the Mn. We only have the O left to fix. This is easy and we'll have 5/2 in front of the O₂ ...

Of course, we don't like fractions and we'll end up multiplying everything by 2 to get rid of the 5/2.

We end up with

4, 2, 4, 5 as coefficients.

Hopefully this examples shows the math required to balance. After all, balancing is simply the law of conservation and understanding that chemical reactions are rearrangements of atoms and not creating or destroying them.

Hope this helps.

[AWK]

The reaction given on the picture is not a redox reaction

[Borek]

http://www.chembuddy.com/?left=balancing-stoichiometry&right=toc

http://www.chembuddy.com/?left=balancing-stoichiometry&right=algebraic-method

[cathy75]

Try to practice and research more on how to solve it. mailbox posts