Before you can start doing full blown reaction stoichiometry problems you're going to need a fully balanced chemical reaction. A balanced chemical reaction will show the correct ratios (in moles) of reactants and products. The big underlying concept here is conservation of mass. This is a fancy way of saying that you have to have exactly as many atoms out (products) as you have coming in (reactants). This is straight up chemical accounting work. You will need to total up each of the different kinds of atoms (elements) on the left side of the arrow and make sure that you get that same exact total on the right side of the arrow. This is what we mean by balancing the equation or reaction. Below is an example of a properly balanced equation.

The whole numbers used to balance each chemical formula are known as **coefficients**. There are actually "four" coefficients here because anytime a coefficient of 1 (one) is needed, it is just left blank for clarity. We know that if a formula is given, there has to be at least one of that formula. Always remember to count the ones when asked about coefficients. Any number other than one must be explicitly shown. We typically want the balanced equation to be balanced with whole numbers. And those whole numbers should be the smallest set of whole numbers that correctly convey the ratios. What is wrong with this version of the previous reaction?

3 CH_{4} + 12 Cl_{2} → 3 CCl_{4} + 12 HCl

Even though there are the same number of atoms on the left and right sides of the equation, those ratios: 3/12, 3/3, 12/12 can all be reduced to a smaller set of numbers with the same ratios. All those coefficients can be divided by 3 and get a new whole number. So this version of the reaction above is not correctly balanced according to accepted norms in chemistry.

The first type of balancing problems that you will complete are done **by inspection**. Balancing by inspection is just that... you look on each side of the arrow and put in coefficients so as to balance out a particular element. Then you do it again. And again. As you go back and forth you will eventually arrive at the proper set of coefficients that perfectly balance all the atoms in the reaction.

So no specific steps here, although there are some pretty good rules of thumb for balancing by inspection. First, try to start with elements that are only in one place (formula) on each side of the equation. Elements that show up numerous times should always be tackled last when balancing. By doing this, you'll most likely have the reaction balanced before having to even attempt the multilisted elements.

Here is a rather classic balancing example of a combustion reaction. In this case, it will be the combustion of ethanol. The unbalanced equation is

C_{2}H_{6}O + O_{2} → CO_{2} + H_{2}O

**Step 1: ** First thing is to notice that carbon (C) is only in one formula on each side of the reaction arrow. So carbons can be easily balanced by having a 1 for a coefficient on the left side (reactants) and a 2 on the right side (products):

1C_{2}H_{6}O + O_{2} → 2 CO_{2} + H_{2}O

**Step 2: ** Now we see that hydrogen (H) is now set at 6 on the left. So we balance the H's with a 3 in front of the water formula:

1C_{2}H_{6}O + O_{2} → 2 CO_{2} + 3 H_{2}O

**Step 3: ** Now we can finish by counting all the oxygens on the right side (we get 7) and adjusting the coefficient for the oxygen (O_{2}) so that the left side also has a total of 7. Note there is one oxygen in the formula for ethanol, this means we only need 6 more which means a coefficient of 3 for O_{2}:

1C_{2}H_{6}O + 3 O_{2} → 2 CO_{2} + 3 H_{2}O

This is the correctly balanced equation shown above. However, we typically do not bother writing/showing coefficients of one because that is assumed if there is a formula there in the first place. So here is the more "streamlined" reaction balanced:

C_{2}H_{6}O + 3 O_{2} → 2 CO_{2} + 3 H_{2}O

I'd go on and on about balancing equations but you could just read up on it from OpenStax. Here is the OpenStax link about Writing and Balancing Chemical Equations.