Balancing Equations?

Balancing equations can be a lot simpler than it seems. What you want to do, is look at each side and see what individual elements make it up. Then, you want to make a table, with the reactants on one side and the products on the other, and list how many of each element are present. Here's an example for C3H8+O2-->CO3+H2O

C 3 - C 1

H 8 - H 2

O 2 - O 4

Once you set this up, you need to try to balance. For example, if you put a 3 infront of CO3, then you would get 3 C's (like you need to equal the reactants) and 9 O's, plus the one O from H2O. Next, balance the H's. 4*H2=8H. However, that means 4 more O's. Now we have 3 C's, 13 O's and 8 H's

C 3 - C 3

H 8 - H 8

O 2 - O 13

Now, see what is not balanced. The oxygen is not, so we need to make it close. However, because it is O2, we will only be able to get an odd number. We have even more work ahead of us. What if we put a 2 infront of the C3H8? Then we balance again

C 6 - C 6

H 16 - H 16

O 34 (if I do 17 O2's) - O 34 (6*3)+(16*1)

It is a little bit of guess and check...but here is the final balanced equation: 2C3H8 + 17O2 --> 6CO3 + 16H2O

C3H8 + O2 ---> CO3 + H20

An easy way of balancing is using fractions.

First balance C, put 3 on the right side.

Balance H, put 4 on H2O

Lastly, count how many oxygens are on the right side, we have 9 + 4 = 13

so on the left side put 13/2. As you cannot leave any fractions on a chemical equation, multiply everything by 2

So we get: 2C3H8+13O2 ---> 6CO3 + 8H2O

C3H8 + O2 --> CO3 + H2O

Look at carbon, there are 3 on the left side so lets make 3 on the right side. Also look at hydrogen and balance.

C3H8 + O2 --> 3CO3 + 4H2O

Now on the left side there are 3 carbons, 8 Hydrogens, and 2 oxygen. On the right there are 3 carbon, 8 Hydrogen, and 13 oxygen.

Now we need to make the number of oxygen even on the right

side. So change 3CO3 to 6CO3.

2C3H8 + 12O2 --> 6CO3 + 8H2O

Hope this helps, it is easier to explain in person.

It doesn't matter if there are two separate oxygens, just make sure that each atom matches up on each side. So for the C3H8+O2 -->CO3+H2O, you have 2 oxygens on the left and 4 on the right, 3 carbons on the left and 1 on the right, and 8 hydrogens on the right and 2 on the left. You just have to keep messing around with the numbers until you have equal amounts of each atom on each side. So for this equation you would end up with 2C3H8 +13CO2-->6CO2 + 8H2O.

To balance a chemical equation you need to get the same number of atoms on each side of the arrow, it's ok if they are broken up between more than one molecule as long as the total on the left equals the total on the right

for C3H8 + O2----->CO3 + H2O

C3H8 + 5O2------>3CO3 + 4H2O

each side has 3 carbons, 8 hydrogens, and 10 oxygens

Hope that helps

******Well first off, are you sure that it's not CO2 instead of CO3 right after the arrow, because whenever you have CH's + O2 it makes CO2 and H2O---its a combustion reaction.******

If it is C3H8 +O2 ---> CO2 + H2O:

1) Easy thing to to is to make the carbons on both side equal---- so it would be 2C3H8 and 6CO2---- you have 6 carbons total on both sides.

2) Next make the hydrogens equal----so since you now have 2C3H8 (which equals 16 H's total on the left side of the arrow) you need 8H2O--- now you have 16 H's on both sides.

3) Last do the oxygens, it's kinda tricky--- now that you have 6CO2 (12 O's) and 8H2O (8 O's) you have 20 O's all together on the right side of the arrow. Since the molecule on the left side is O2 all you need is 10O2 to equal 20.

And you are done!!!

So in all you should have:

2C3H8 + 10O2 ----> 6CO2 + 8H2O

ok here's what i would do to balance the equation...

first balance out the carbons... by putting a 3 infront of CO3...

then balance out the hydrogens... by putting a 4 in front of H20...

at this point you'd have C3H8+O2-->3CO3+4H20...

however the O's still aren't balanced... on the right you have 13 O's... and on the left you only have 2...so then you find the number that you would multiply by 2 to get 13 ----> 6.5...

so at this point you have C3H8+6.5O2--> 3CO3+4H2O...

now you can't have a fraction as one of your coefficients... so now you multiply all the coefficients in the equation by 2 to fix that fraction...

giving you:

2C3H8+13O2-->6CO3+8H20

hope that helps

There is a easy method to balance complex eqns. like the one stated

C3H8 + O2 -->CO2 +H2O

Follow the given steps

1. First write the names of the elements present and the no. of times they appear in the eqn.

i.e, C appears 2 times

H 2 times

O 3 times

2. start balancing with the element which appears least no. of times. If two elements appear same m\no. of times, balance the element with highest 'Z' first.

In the given eqn. order of balancing is C , H , O

To balance C keep 2 before CO2

Then the eqn. becomes C3H8 + O2 --> 2 CO2 + 4 H2O

To balance O of which RHS has 8 atoms keep 4 before O2 in LHS.

Thus eqn. becomes C3H8 + 4 O2 --> 2 CO2 + 4 H2O

I THINK i AM RIGHT .. in ur question u gave product as CO3 WHICH DOESNOT EXIST.. i THINK it is CO2

If u first tryed C3H8 + O2 ------> 3CO3 + 4H20 u were on the right track but ur oxygen O2 could be made into 13

so you make the numbers larger

answer

2C3H8 + 13O2 ---> 6CO3 + 8H2O

SRY not good at explaining

by the way im right 100percent sure

32rr4t