what are the steps to balencing an equation? Please make it as easy as possible I'm a mom and in my 40s I don't remember doing this in school!

Mathematically speaking, there are all manners of ways to balance and equation.

A balanced equation simply means that the left side of the "=" sign is the same as the right side. For example, 3 = 3.

Can you supply an actual equation that is giving you problems to us.

I do not have one at this time but when my girl gets home ill put them on , thank you for your quick response its greatly appreciated.:)

Dear lady,

Is your little lady having problems with math or chemistry?

Math I can help with. Chemistry is a totally different story!

what are the steps to balencing an equation?

Usually, I balance the metal(s) first, then the anions (negative species). If hydrogen and oxygen are involved, balance the oxygen first -- usually by multiplying waters, and then the hydrogen.

In redox (reduction / oxidation) reactions, I balance them using "half reactions". You balance the oxidizing agents and the reducing agents separately using electrons. You then multiply the two equations appropriately (electrons on one will be on the left and on the right on the other) and add them together. You'll have the same number of electrons on the left and on the right, so you can eliminate them (if what goes in also comes out, it's not involved in the reaction).

That's pretty general. I can do it every time, but it's hard to tell exactly what I do to accomplish it.

Usually, I balance the metal(s) first, then the anions (negative species). If hydrogen and oxygen are involved, balance the oxygen first -- usually by multiplying waters, and then the hydrogen.

In redox (reduction / oxidation) reactions, I balance them using "half reactions". You balance the oxidizing agents and the reducing agents separately using electrons. You then multiply the two equations appropriately (electrons on one will be on the left and on the right on the other) and add them together. You'll have the same number of electrons on the left and on the right, so you can eliminate them (if what goes in also comes out, it's not involved in the reaction).

That's pretty general. I can do it every time, but it's hard to tell exactly what I do to accomplish it.

Can you please give me an example

Balancing half-reactions (oxidation-reduction only)
Half-reactions are fairly easy to balance. Consider this reaction:

Fe^{2+} \rightarrow Fe^{3+} + e^-

This is the oxidation of iron. It's simple because there's nothing involved except an electron and an iron atom. It's not actually meaningful in any real sense until we have an oxidizing agent. Here's the reduction of chlorate to chloride. (whether it really goes this far is irrelevant to the discussion. We're simply balancing the equation).

ClO_3^- + e^- \rightarrow Cl^-

This half-reaction needs more balancing. Balancing a half reaction usually involves knowing whether it's in acid or basic solution. However, you can always balance it as if it were in acid solution, and then correct it for basic solution. I'll show you how that's done. I'm going to delete the electron for starters. We'll add it in later.

Step 1, balance the principal element. In this case, it's the chlorine atom that's being reduced. In this case, it's trivial because there's one chlorine atom on the left and one on the right. I won't even bother to rewrite it.

Step 2, Balance oxygens by adding water

ClO_3^- \rightarrow Cl^- + 3H_2O

Now there are three oxygen atoms on the left and three on the right.
Step 3, Balance hydrogen atoms by adding H+

ClO_3^- +6H^+ \rightarrow Cl^- + 3H_2O

Step 4, Balance charge by adding electrons

ClO_3^- +6H^+ +6e^- \rightarrow Cl^- + 3H_2O

We now have two balance half reactions. Multiply one or both of the reactions by some number so that we have the same number of electrons on one side as the other. A fully-balanced reaction shows no electrons. Electrons are only present in half-reactions. Note that electrons must be on opposite sides of the half reactions. In other words, one half-reaction has to be an oxidation and the other must be a reduction.

6 \ times \, (Fe^{2+} \rightarrow Fe^{3+} + e^-)

6 Fe^{2+} \rightarrow 6 Fe^{3+} + 6 e^-)

ClO_3^- +6H^+ +6e^- \rightarrow Cl^- + 3H_2O

Now add the half reactions and delete things on both sides that are the same. If water or H+, for example, were on both sides, eliminate some from one side and decrease by the same amount on the other. In this case, only e- is seen on both sides.

6Fe^{2+} + ClO_3^- +6H^+ \rightarrow 6Fe^{3+} + Cl^- + 3H_2O

You can add spectator ions, if you wish. Make sure you add the same ions to both sides of the equation (they are, "spectator ions", after all).

6FeCl_2 + KClO_3 +6HCl+ \rightarrow 6FeCl_3 + KCl + 3H_2O

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If you're in basic solution, use the water equilibrium to change from H+ to OH- when you're still in the half-reaction mode.

H^+ + OH^- \rightleftharpoons H_2O

ClO_3^- +6H^+ + 6OH^- +6e^- \rightarrow Cl^- + 3H_2O + 6OH^-

ClO_3^- +6H_2O +6e^- \rightarrow Cl^- + 3H_2O + 6OH^-

ClO_3^- +3H_2O +6e^- \rightarrow Cl^- + 6OH^-