# Introductory of Chemistry

1. Oct 14, 2007

### leah0084

1. The problem statement, all variables and given/known data

Balance the following equations. (Use the lowest possible coefficients and enter a "1" if the coefficent is a one.)
(a) ___Cr(s) + ___S8(s) -> ___Cr2S3(s)
(b) ___NaHCO3(s) -> ___Na2CO3(s) + ___CO2(g) + ___H2O(g)
(c) ___KClO3(s) -> ___ KCl(s) + ___O2(g)
(d) ___Eu(s) + ___HF(g)-> ___EuF3(s) + ___H2(g)

2. Relevant equations

3. The attempt at a solution

i just do not know how to do this.
could anyone simply explain how to do?
once i understand the basic formula to solve one problem, i should be able to do the rest.

Last edited: Oct 14, 2007
2. Oct 14, 2007

### bob1182006

you want to have the atoms on one side be equal to the number of atoms on the other side.

for a you should get the S's to be equal first since Cr is by itself so you can balance that out really easy.

So what numbers can you multiply by to get S8 and S3 to have the same number of S atoms?
and when you have those balanced how many Cr do you have on the right side? multiply Cr on the left by that number and it's balanced.

With the others it's a bit more complicated, usually it's a hit and miss sort of thing, first balance out this and oops doesn't work go back and try again :s....

3. Oct 14, 2007

### ace123

Basically you want both sides of equation to be equal. For example H2 +O2-> H2O. To balance both sides we need 2 oxygen on left and 2 oxygen on right side. We can not change subscripts so we change the coeffecient in front of H2O to a 2. Therefore you have 4 hydrogen on right but only 2 on left side. So we change the coeffecient in front of H2 to 2. So in the end we have 2H2+O2-> 2H2O. Same concept for your problems just have different elements and compounds. Hope you understand