Question

Balance each of the following equations

Answer 1

`2Sb+3I_2=2SbI_3`

`P_4+3O_2=2P_2O_3`

`2SO_3=2S+3O_2`

`2Fe_2O_3+3C=4Fe+3CO_2`

`CH_4+2O_2=CO_2+2H_2O`

Explanation:

Each side of the equations need to have the same number of each element.

Equation 1

Since there are 2 I on one side of equation 1, and 3 on the other side, I will do the following:

`Sb+3I_2=2SbI_3`

Now the number of Sb is unbalanced, so I balance it like so:

`2Sb+3I_2=2SbI_3`

Equation 2

Using the same strategy, I focus on one element at a time. I can balance the P like this:

`P_4+O_2=2P_2O_3`

Now, I can balance the O. There are 6 on the right, so if I put 3 on the left, there will be 6 there as well!

`P_4+3O_2=2P_2O_3`

Equation 3

Starting with O this time, I put 2 on the left and 3 on the right to bring them both to 6.

`2SO_3=S+3O_2`

Now there are 2S on the left and only 1 on the right, so I balance it:

`2SO_3=2S+3O_2`

Equation 4

Starting with O again, there are 3 on the left and 2 on the right.

`2Fe_2O_3+C=Fe+3CO_2`

Now, the Fe are unbalanced as well. There are 4 on the left and 1 on the right:

`2Fe_2O_3+C=4Fe+3CO_2`

Finally, balance the C, there is 1 on the left and 3 on the right.

`2Fe_2O_3+3C=4Fe+3CO_2`

Equation 5

This one is pretty hard to read, so I will assume that the equation looks like this (please comment if this is incorrect)

`CH_4+O_2=CO_2+H_2O`

Let's balance the H first:

`CH_4+O_2=CO_2+2H_2O`

Now the O. There are 2 on the left and 4 on the right.

`CH_4+2O_2=CO_2+2H_2O`