Balancing equations

[cosmic-chris]

Hi can't remember how to balance chemical equations,can any one shed some light on the easiest method for this?

[insane_alien]

count up the atoms on each side of the equation, make sure they're equal.

 

thats about it. it becomes intuition after a while.

[cosmic-chris]

So H2 + O2 ----> H20 why are we missing one oxygen?or am i missing somthing ?

[insane_alien]

well you need to compensate. you need 2 O's on the right hand side so multiply that side by 2. then you'll need 2 mor H's on the left side so multiply that part by 2.

[cosmic-chris]

Methane gas reaction

 

CH4 + O2 -------> CO2 + H2O

 

As there are four oxygen on the right side of the equation and four hydrogen on the left we need to balance that out so it would become,

 

CH4 + 2 O2 ------> CO2 + 2 H2O Would this equation be correct if noy let me know where i am going wrong.

[insane_alien]

looks good.

[cosmic-chris]

And the water equation would be 2 H2 + O2 -----> 2 H2O is that correct?

 

I think i have got the hang of it again thanks insane_alien

[insane_alien]

yeah, its all good.

[thedarkshade]

If you get to more tricky equations, I'd suggest you to use oxidation numbers. You figure out the changes in oxidation states and then place those number in the molecules in the equation. Watch:

 

[ce]KMnO4 + H2O2 + H2SO4 -> K2SO4 + MnSO4 + O2 + H2O[/ce]

 

it's kinda hard to balance without looking at the oxidation number in this case. The changes that happen here are:

[ce]Mn^{+7} ->Mn^{+2}[/ce] so it takes 5 electrons

[ce]O^{-1} -> O^{0}[/ce] so it loses 1 but since O is a double-atomic molecule we use 2. Then we just change the places and now where we have Mn we put 2 and where we have O we put 5, se we have:

 

[ce]2KMnO4 + 5H2O2 + 3H2SO4 -> K2SO4 + 2MnSO4 + 5O2 + 8H2O[/ce]

 

It's a bit tricky but very useful!]

 

Cheers,

Shade