[eeeeeeeee]

here are a few examples

Na +I2 (arrow going right) NaI


N2 + O2 (arrow going right) N2O

I dont understand how to do this. and none of my class mates do eather.

[mrl224]

The easiest way to do this is by counting up the total number of atoms you have on each side.  Then you can change the stiochemetric coefficient infront of each of the compounds so that the number of atoms are the same on both sides.

E.g. Your first question: Na + I2 -> NaI
On the reactants side you have 1 atom of sodium and 2 atoms of iodine, on the products you have 1 atom of sodium and 1 atom of iodine.  To correct this imbalance you must place a coefficient of 2 infront of NaI and a 2 infront of Na so you have:
2Na + I2 -> 2NaI
2 atoms of sodium in reactants, 2 atoms of iodine in reactants, 2 atoms of sodium in products and 2 atoms of iodine in products.  These problems just take a lot of practice.

[eeeeeeeee]

Thanks. I see how you would do one like that but what about ones like these

S+02-- SO3

and
 CH4+ O2-- CO2 + H20

these ones are diffrent then the others

[xiankai]

try balancing one element at a time, then move on to the next (in any order u like)

for eg.

H2O2 --> H2O + O2

balance say, H first. but H is already balanced, so we move on to O

2H2O2 --> 2H2O + O2

these are quite easy when u get used to it. try those 2 yourself.

[eeeeeeeee]

WOW. Thanks i understand how to do small ones. but no a big one like this.

(i know it seems like your doing my homework for me but your not)


P4O10+ H2O---H3PO4

[mike]

You have four P atoms on the left hand side of the equation so you need to have four P atoms on the right hand side of the equation in order fo it to balance, right?


P₄O₁₀ + H₂O ---> H₃PO₄

so put a four in front of H₃PO₄ to give you four P atoms:


P₄O₁₀ + H₂O ---> 4 H₃PO₄

You can add up the rest of the atoms on the left hand side and see if they balance with the atoms on the right hand side.