let's do some practice problems for balancing Chemical Equations we'll start off with some examples that are pretty basic and straightforward and then the problems will get more challenging as we move on here's our first equation we got Xenon and Florine we want to keep track of how many atoms of these elements we have on both sides of the equation so we're going to make a little chart okay I got Xenon and Florine on this side and then Xenon and Florine on this side over here I have one Xenon atom one and then I have F2 so I got two flines over here also one Xenon and then F6 so I got six flines now take a look at these numbers this equation isn't balanced yet because we have different numbers of the atoms for one of the elements I've got two Florine here but I've got six Florine here so it's not balanced in in order to balance it I've got 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 here's how I'm going to do it I got six Florine here but I got two Florine here so for this to balance I need more flines on the left side I can add a number in front of the F2 here if I put three in front front of this F2 here we'll put it in I put three in front of this F2 now I have three * 2 gives me six Florine and now they balance one one for the xenons six and six for the Florine so now this is a balanced equation by adding this number by adding this coefficient now really quick this is one very common question people often ask wait why did you put that three there couldn't you just change this two to a six and then it would also balance no no no no no no no no no you can't do that super common misconception 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 you can do is you can put numbers in front of the elements or compounds okay but you can't change or add the subscripts okay so that's how you balance an equation let's do a whole bunch more practice okay so in this equation we've got three elements silver AG hydrogen H and sulfur s obviously we've got these on both sides of the equation okay so let's see how many atoms we have of each over here we got one AG we got H2 so we got two two hydrogen atoms and then we have S so we got one sulfur atom over here we have ag2 so we have two Silvers here um uh sulfur we have one and then we have H2 so we have two hydrogens okay what balances and what doesn't well the hydrogen's and sulfur both balance but the 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 just put a two in front of this AG so now instead of having one AG now I have two and now I got two two and one it balances this equation is a little bit more challenging because we have oxygen in all the compounds here that means we're just going to have to be a little bit more careful when we're adding up the number of oxygen atoms that we have okay so the elements in this equation we got k potassium O oxygen H hydrogen and C carbon and over on this side we've got pottassium oxygen hydrogen and carbon okay how many of each of these do we have okay we got one potassium here now for oxygen this is where you just got to be a little bit more careful okay because we have one oxygen here but then we have two oxygens here so 1 + 2 total we have three then hydrogen we got one of those and carbon we got one of those over here for pottassium we have two of these because we got K2 then uh oxygen we have three there and we have one there it's going to give us four total then we have H2 we have two hydrogens and one carbon okay so what balances what doesn't the carbon's balance but other than that we've got to add some coefficients to change this stuff around okay oxygen doesn't balance but since oxygen is in every one of the compounds I'm going to leave that for later I'm going to start with potassium because it's easier right I have two potassiums here but I've only got one here so the first thing I'm going to do is I'm going to put a two in front of this Koh and let's see what that does okay the first thing that's going to do is it's going to give me two and instead of one potassiums because I have the two * K now for oxygens let's look at how it's going to change that now I'm going to have two oxygens because two times that plus the two that I have over there in CO2 so I'm going to have 2 + 2 is now going to give me four so I've got four oxygens and in terms of hydrogen now I'm going to have two * one hydrogen so I'm going to get two hydrogens and check it out putting this putting this two here changed everything so so now I've got 2 4 2 1 it all balances now we're going to start doing some equations that require more than one step to balance okay so this one here has sodium and chlorine over here we got one sodium and we have cl2 we have two chlorine atoms over here we have na so we have one sodium one Na and one cl okay how am I going to start well I have more chlorine on this side I have two over here and I only have one here okay so I can start out by putting a two in front of Na okay that's going to give me 2 CL but look what else it's going to do okay now I have two * na so I have two Na's I also have two * CL okay so I also have two cl's so now the cl's balance but now I have two Naas two sodium on this side and only have one on this side so now to balance this out I'm going to put a two in front of this na here and now I have two of these so now on both sides of the equation I have two sodiums and two chlorines the equation balances okay iron oxygen and carbon over here I have one iron one oxygen and one carbon here I have one iron two oxygens and one carbon okay how am I going to balance this the first thing I'm going to do is I'm going to balance these oxygens okay so I have two over here and I have one here I'm going to change that by putting a two in front of the feo okay so that's going to give me two oxygens but it's also now going to multiply this Fe by two so now I'm going to have 2 Fe okay so now I've got 2 Fe here and one there so now these don't balance so the next thing I'm going to do is I'm going to put a two in front of this Fe there so now instead of having one I have two of them now I have two two one it balances okay let's look what we have here we have silicon on this side we have one of them O2 two oxygen and one carbon over here we have one silicon oxygen we have one and carbon we have 1 + one so we have two okay so what doesn't balance well the carbon don't balance and the oxygen don't balance I'm going to leave the carbon alone for now because it's in both of these compounds so I I kind of don't want to mess with it right now but I could I'm just going to start out with the oxygen okay I have two oxygen on this side and I have one oxygen on this side so I'm going to multiply the co by two so that can give me two oxygens but now it's also going to do something else right because we're also multiplying the C by that two so that means means now for the total carbons I'm going to have the one from there plus now the two from there and that's going to give me a total now of three okay so now I have three carbons here I've got one carbon here so I'm going to put a three in front of this carbon that's the only Carbon on this side so now I have 3 * 1 is 3 1 2 and 3 and they balance this equation has five different elements in it and it also has these parentheses so we'll talk about these in a minute first the number of atoms iron Fe we have one here cl3 na1 o and H we have one of those okay now Fe over here we have one and then we got the parentheses okay so the parentheses mean that everything inside here is multiplied by three okay so we have o * 3 which means we have 3 O's and we have H * 3 so that's three H's and then we have Na and Cl so one each of those okay so where are we going to start with this well I have uh I have an imbalance in my oxygens here so I might as well just start there okay so three here one over here so I'm going to put a three in front of NaOH okay so that is going to multiply all the stuff by three I'm going have 3 na 3 o and 3 H okay what did that do for me well that Balan the oxygens and hydrogens but now the sodium the na I have three on this side and I only have one on this side okay so I can fix that by multiplying This na CL by three here okay so that's going to give me 3 Na and it's also going to be three times a CL so now I have 3 CL and check it out that fixed it I have three CL over here but I got three CL there great so 1 3 3 3 3 it balances we got some more parentheses here let's take a look at this equation okay I have one aluminum two hydrogens one sulfur and four oxygens then over here I have two aluminums everything in here is multiplied by three so that means that I have three sulfur and I have 4 * 3 which is 12 oxyg and then I have two hydrogens okay what am I going to do first well take a look at we have aluminum and we have hydrogen on their own so I'm going to save them for last I don't want to balance them right now because we can use these at the end to to straighten out some of these details so let's look right now at sulfur and oxygen I can fix both the sulfur and the oxygen and balance them by multiplying these by three right I have three sulfurs and 12 oxygen so if I just multiply these by three I can get them to balance so I've got sulfur and oxygen there so I'm going to start by putting a three in front of that okay that's then going to give me three sulfurs and 3 * 4 12 oxygen and it's also going to give me some more hydrogens okay so now I got six hydrogens okay so we got the sulfur and the oxygen to balance now just by multiplying by three now my hydrogen my hydrogen doesn't balance I have six on this side two on this side so I can fix that by mult multiplying the hydrogen by three now I have 3 * 2 on this side which is going to give me six and then take a look at the aluminum I have two on this side one on this side but I can fix that by multiplying by two over here and then I get those to balance and everything's good you may have noticed these are getting a little bit more complex it takes more and more steps to balance them okay so for this we have one nitrogen we have three hydrogen we have one CU which is copper and we have one oxygen over here we have two nitrogens uh two hydrogens uh one copper and one oxygen okay where am I going to begin with this well copper and nitrogen are on their own here so I'm going to leave them to last I'm going to focus in on hydrogen and oxygen the oxygen are balanced right now but the hydrogens are not and check out I've got this 32 thing going on I can solve that by doing a crisscross multiplication to get six so the hydrogens are here I'm going to multiply this by two and then I'm going to multiply this by three okay what that's going to do is it's going to give me 2 * 3 six hydrogens it's also going to give me two nitrogens and then over here it's going to give me 3 * 2 hydrogen which which is 6 and then 3 * 1 which which is three oxygens okay so now I've got my hydrogens balanced the next thing that I can do is um leaving copper and nitrogen for later let's balance the oxygens okay I have three on this side I have one on this side my oxygen is there so I'm going to multiply this by three so I'm going to do 3 * 1 get three oxygens but I'm also going to get three * 1 for copper okay so now my Oxygen's balanced but my copper my copper is still a problem here okay so what I'm going to do there is I need three on this side because I have three on that side but luckily copper is by itself so that's what I say like at the very end you can go through and and fix this stuff so I'll put a three there in front of copper so 3 * 1 gives me three for copper so this balances now