Balancing Chemical Equations Practice Problems

Let's do some

practice

problems

for

balancing

chemical

equations

. We'll start with some examples that are pretty basic and simple and then the

problems

will get more challenging as we go. Here is our first equation. We have xenon and fluorine and we want to track them. how many atoms of these elements we have on both sides of the equation, so let's make a little chart. Well, I have xenon and fluorine on this side and then xenon and fluorine on this side. Here I have a xenon atom, one and then I have f2, so I have two fluorine here also a xenon and then f6, so I have 6 fluorine.

Now look at these numbers, this equation is still not balanced because we have different numbers of atoms for one of the elements i. I have two fluorine here but I have six fluorine here so it is not balanced, to balance it I have to add numbers or coefficients in front of one or more of these elements and compounds to change the number of atoms that I have on the different sides of the equation , okay, this is how I'm going to do it. I have six fluorine here but I have two fluorine here, so for this to balance out I need more fluorine on the left side.

I can add a number in front of the f2 here if I put 3 in front of this f2 here we put it I put 3 in front of this f2 now I have three times two it gives me six fluorine and now they balance one one for the Xenon ones 6 and 6 for the fluorine , so now this is a balanced equation adding this number adding this coefficient now real quick this is a very common question that people often ask wait why did you put that 3 in there couldn't you just change this 2 to a 6 and then it would also balance no no no no no no no no no you can't make that very common mistake you can't change these subscripts here you can't change them you can't add them so you can't change this to a six you can't put a six here or anything like that, the only thing What you can do is put numbers in front of the elements or compounds, okay, but you can't change or add the subscripts, this is how you balance an equation, let's do a lot more

practice

, okay, so in this equation we have three elements silver AG hydrogen H and sulfur s, obviously we have them on both sides of the equation, okay, so let's see how many atoms we have of each. here we have one AG we have h2 so we have two hydrogen atoms and we have s so we have one sulfur atom here we have AG two so we have two silvers here on sulfur we have one and then we have h2 so we have two hydrogens okay what balances and what not, both hydrogen and sulfur balance each other, but silver we have two here and one here, we can fix it by adding a number or coefficient in front of one or more of these elements and compounds.

I need more. silvers on this side and luckily I can put a two in front of this AG, so now instead of having one AG I now have two and now I have two, two and one. Balancing this equation is a little more challenging because we have oxygen. in all the compounds here that means we're going to have to be a little more careful when we add up the number of oxygen atoms that we have, so the elements in this equation we have K potassium Oh oxygen H hydrogen and Carbon C and on this side we have potassium, oxygen, hydrogen and carbon.

Okay, how many of each of these do we have? Okay, we have a potassium here now for oxygen. This is where you need to be a little more careful. Okay, because we have one oxygen here, but then we have two oxygens here, so one plus two total we have three, then hydrogen we have one of those and carbon we have one of those here for potassium, we have two of these because we have k2 and then oxygen. we have three there and we have one there it will give us four total then we have h2 we have two hydrogens and one carbon okay so what balances what doesn't balance the carbons but other than that we have to add some coefficients to change this, okay, the oxygen does not balance, but since the oxygen is in each of the compounds, I will leave it for later.

I'm going to start with potassium because it's easier. I have two potassium here, but I have I only have one here, so the first thing I'm going to do is put a two in front of this Koh and let's see what it does. Well, the first thing I'm going to do is give me two instead. of one potassium because I have twice that K now four oxygens we are going to see how that is going to change now I am going to have two oxygens because twice that plus the two that I have there in co2 so I am going to have two plus two now it is going to give me four so I have four oxygens and in terms of hydrogen now I'm going to have two hydrants for one.

I'm going to get two hydrogens and check it by putting this. These two here changed everything, so now I have two four two one, everything balances out. Now let's start doing some

equations

that require more than one step to balance. Well, this one here has sodium and chlorine. Here we have one. sodium and we have CL two we have two chlorine atoms here we have NaCl so we have one of sodium one of na and one of Cl okay, how am I going to start? I have more chlorine on this side, I have two here and I only have one here, okay, so I can start by putting it two in front of na, look, oh, okay, it's going to be two CL, but look what else it's going to do.

Now I have two n a times, so I have two n A. I also have two. times CL, okay, so I also have two CLS, so now the CLS is balanced, but now I have two n A for the sodium on this side and I only have one on this side, so now to balance this, I'm going to put it Not in front of this, here and now I have two of these, so now on both sides of the equation I have two sodium and two chlorine, the equation balances iron, oxygen and carbon well. Here I have an iron, an oxygen and a carbon.

Here I have a two iron. oxygens and a carbon, okay, how am I going to balance this? The first thing I'm going to do is balance these oxygens, okay, so I have two here and I have one here. I'm going to change that by putting a 2 in front of the f EO, okay that will give me two oxygens, but now it will also multiply this Fe by 2, so now I will have 2 Fe, ok, now I have 2 Fe here and one there, so now they don't balance out, so the next thing I'm going to do is put a 2 in front of this Fe there, so now instead of having one, I have two, now I have two. one balances ok let's look at what we have here we have silicon on this side we have one of them or two oxygen and a carbon here we have a silicon oxygen we have one and carbon we have one plus one so we have two , ok So what doesn't balance well is carbon and oxygen.

I'm going to leave the carbon alone for now because it's in both compounds, so I don't want to mess with that. Right now, but it could. I'm going to start with oxygen. Okay, so I have two oxygens on this side and I have one oxygen on this side, so I'm going to multiply the EC o by 2 so I can give myself two. oxygens, but now it's also going to do something else right because we're also multiplying this C by that too, which means that now for the total carbons I'm going to have the one over there plus now the two over there and that's going to give Now I have a total of three, so now I have three carbons here.

I have a carbon here so I'm going to put a 3 in front of this carbon which is the only carbon on this side so now I have 3 times 1. 3 1 2 and 3 and they balance this equation it has 5 different elements and it also has these parentheses, so we'll talk about these in a minute first, the number of iron atoms Fe, we have one here CL 3 na 1 o and H we have one of those, okay, now Fe here we have 1 and then we got the parentheses, it's okay, so the parenthesis means that everything inside here is multiplied by 3, okay, so we have Oh multiplied by 3, which means we have 3 OS and we have H multiplied 3, so that's three H and then we have Na and Cl, so one of each of them, well, where do we start with us?

Well, I have an imbalance in my oxygens here, so I might as well start there, okay, so three here. one over here, so I'm going to put a 3 in front of NaOH, okay, that's going to multiply all things by three. I have three na 3 o + 3 H, ok what does that do for me? Well, that balances the oxygens and hydrogen, but now the sodium na I have three on this side I don't have one on this side, okay, so I can fix that by multiplying this NaCl by three here, okay, that will give me three na and it will also be three multiplied by CL, so now I have three CL and look, that fixed it.

I have 3 CL here but I have 3 CL there. Great, so 1 3 3 3 3 balances out, we have more parentheses here, let's take a look at this equation. Well, I have one aluminum, two hydrogens, one sulfur and four oxygens. So here I have two aluminums. Everything here is multiplied by three, which means I have three sulfurs and four times three, which is 12 oxygens and then I have two hydrogens. Well, what am I going to do first? Take a look at this. We have aluminum and we have hydrogen on their own, so I'll save those for last. I don't want to balance them now because we can use them in The end is to clarify some of these details, so let's see now what is sulfur and oxygen.

I can fix both the sulfur and oxygen and balance them by multiplying them by three. I have three sulfur and twelve oxygen, so if I just multiply them by three I can get them to balance, so I have sulfur and oxygen in there, so I'll start by putting a 3 in front of that, okay, that will then give me three sulfur and three times four twelve oxygens and it also goes to give me more hydrogens, okay, now I have six hydrogens, okay, so we balance the sulfur and the oxygen now by simply multiplying by three. Now my hydrogen, my hydrogen is not balanced.

I have six on this side, two on this side. I can fix that by multiplying the hydrogen by three. Now I have three times two on this side, which will give me six and then look at the aluminum. I have two on this side and one on this side, but I can fix that by multiplying the hydrogen by three. multiplying by two here and then I make them balance and everything is fine, you may have noticed that they are getting a little more complex, it takes more and more steps to balance them, okay, so for this we have a nitrogen, we have three hydrogens. a Cu, which is copper and we have an oxygen here, we have two nitrogens, two hydrogens, a copper and an oxygen, well, where am I going to start?

Well, the copper nitrogen is alone here, so I'll leave you. Lastly, I'll focus on hydrogen and oxygen, although the oxygens are balanced at the moment, but the hydrogen is not, and look, I have this 3-2 thing. I can solve it by doing a cross multiplication to get 6 so the hydrogens are here I'm going to multiply this by 2 and then I'm going to multiply this by 3 ok what that's going to do is it's going to give me 2 times 3 6 hydrogens is also going to work for me. to give 2 nitrogen and then here it is going to give me 3 times 2 hydrogens, which is 6 and then 3 times 1, which is 3 oxygens, well, now that I have my hydrogen balanced, the next thing I can do is leave copper and nitrogen. for later let's balance the oxygens, ok, I have 3 on this side, I have one on this side, my oxygen is there, so I'm going to multiply this by 3, so I'm going to do 3 times, I'll get 3 oxygens, but I'll also get 3 times 1 for copper.

Well, now my oxygen is balanced, but my copper is still a problem. So what I'm going to do there is I need 3 on this side because I have 3 on that side. but luckily the copper alone so that's what I say at the end you can go through and fix these things so I'll put a 3 in front of the copper so 3 times 1 gives me 3 for the copper so this balances out now .