Balancing Equations using ERP Method

Okay, let's practice balancing equations using the ERP method. Remember, E stands for element. R for reactant and P for product. So the first thing you want to do is label the reactant side and the product side of your equation. Then you're going to list all the elements you have in your chemical equation. So on number one, we have FE, and we have O now, let's count our atoms. Let's start with iron. How many iron atoms do we have over here? How many effies? We have one. So I put one, and the reactant column, the R column. How many O's do I have? I have two, so I'll put two in the reactant column for the reactant side. Now we'll move on to the product side of the equation and the P column. How many FEs do I have? I have two here. And how many O's do I have? I have three. I look at my numbers and see if they match up. One iron on the reactant side and two on the product. That is not balanced. In order for it to be balanced, they must be equal. So I will write not balanced on the line. All right. Let's look at number two. Okay, again, I'm going to label my reactant side in my product side. And I'm going to list my elements. I have AL. I have H and I have CL. Over here, there's no subscript, which means I only have one. Did I look at my H and there's no subscript, so I have one. I look at CL and there's no subscript, so I have one. Now look at my product side. Find AL. No subscript, which means one. CL has a subscripted three. So I'm going to put a three. And H has a subscript of two. I'm going to put a two. Get another look at my numbers. They do not match. So again, number two is not balanced. Let's look at number three. I'm going to label my reactant side in my product side. I'm going to list my elements. I've got K and BR. Remember, a new element always starts with a capital letter. So in the case of BR, there's a capital B with the lower case R, that's the same element. All right, so how many K's do I have in my reactant side? I have a coefficient of two. And remember, you multiply the two by everything behind it, and that chemical formula. So two times K means I have two K's. However, that too does not carry over to the BR's. That plus sign stops it. So now I'm going to look at my BRs, my bromines. How many do I have? Well, I have a subscript of two here, which means I have two bromines. Now let's look at my product side. I have a coefficient of two, which means I have two KBR molecules. Remember, I multiply that two by everything behind it. So two times K means I have two K's. And two times the BR means I have two BRs. My numbers match. So this equation is balanced. All right, let's look at another one. Keep my page to scroll. All right, I'm going to skip down to number 5. You can do number four on your own. My reactants, I'm going to label my products to my elements. I've got NA. I've got C, I've got O dot H and I have CL. Lots of elements in this one. So I start with my reactants. I have two NA's, because I have a subscript of two there. No subscript by the C so I have one. I have a subscript of three by the oxygen, so I have three. I have a coefficient of two, so I multiply it by everything behind it, two times H means two hydrogens. And two times CO means two chlorines. Let's go to my product side. I have a coefficient of two in front of my NaCL. So two times the NA means two. And two times CL. I'll put a two there. And then I move on to my next, because again, the plus sign moves on to a new formula. I have coefficient of two, so I have two H's. And I have a coefficient of no coefficient there, something to have one O. The coefficient of one there, so I have one C and have a coefficient in there. I have two O's. So I will put plus two equals three I was total. And now I match everything up. Two and two, one and one, three, and three, two, and two. And two and two. Everything matches. So my equation is balanced. Let's look at the next page. Let's try. Well the number 9 together. So I have AL. H and CL. Okay, I'm going to label my reactants and my products. I have a coefficient of two. Times everything behind it, so that's the AL. So I have two ALs. Here I have a coefficient of 6, so 6 times H sorry, 6 times H is 6 H's and 6 times CL is 6 CL's. Now on this side, it gets a little more complicated. I have two times AL. So I have two ales. This one I have two times CL three. I already have three chlorines. I have the coefficient of three. Now I have to multiply that two by that, so it would be two times three chlorine would be a total of 6 chlorine. And when I get down to the next one, three H two. Again, I have two hydrogens here, but I have three molecules of it, so I multiply three times that two for 6 H's. So this one is balanced. Okay, go ahead and complete the rest of them on your own.