let's practice a little bit of balancing here are three different simpler balancing problems for us to start with if you'd like to try them on your own and then go through you could pause the video and then try these and then see whether you get the same answers as i do let's first take a look at the first problem so we have water h2o liquid breaking down into hydrogen gas and oxygen gas first i need to determine the number of elements on each side so i have as it's written i have two hydrogens and one oxygen on the left on the right i have two hydrogens and two oxygens notice our oxygens don't match so i need to have two oxygens to match my two oxygens on the right so i'm going to just put it 2 here for my coefficient so i have two water molecules now that's going to change my oxygens to 2 but it's also going to change my hydrogens to 4. so i have two h2os each one's got two hydrogens so two times two for four hydrogens on the left but so that's on the left that means that i've got four here i only have two on the right but i can fix that by putting a two here as well two times two four hydrogens let's take a look at we always double check two times two four hydrogens two times one so sometimes you might wanna wire it in one two oxygens two times two four hydrogens two oxygens so remember if we don't have a coefficient we assume the coefficient here is one let's take a look at the second question nitrogen gas plus oxygen gas forming n2o5 gas we have two nitrogens two oxygens on the left on the right we have two nitrogens and five oxygens hmm two and five well you can't there's no number that we can put in front of o so in front of the oxygen to make these equal so we need to look at our lowest common multiple so lowest common multiple so what's the lowest number that both two and five go into so two and five so the lowest common multiple of two and five is ten so we're gonna need to have ten oxygens to get 10 oxygens i need to put a 2 in front of here let's get 10 oxygens on the left i'm going to need to put a 5 in for my o2 so my o on the left i'm going to have 10 oxygens on the left on the right i'm also going to have 10 oxygens but i also change my nitrogens here 2 times 2 it's going to give me 4 nitrogens since i have four nitrous on the right i need to have four nitrogens on the left so i need to put a two in front of this one so now i'm going to have four nitrogens on the left ten oxygens on the right 10 oxygens okay so always double check and you check and see is there a lowest common multiple that these values can go into the last example we have hydrogen there's one two sometimes we have elements in more than one place we'll look at different things that we do with this type of compound later chlorine i have one n a i have one an oxygen i have one now let's check my elements on the right side of the equation on my products i have hydrogen i have two hydrogens chlorine i have one chlorine n a i have one n a an oxygen i have one oxygen wait a second two hydrogens two hydrogens one chlorine one chlorine one sodium one sodium one oxygen one oxygen this equation is already balanced the reason i include this is because it's very common for people to jump right into changing coefficients without checking is this actually balanced in this case it is and so there's no more balancing required it's it's correct as written so just to summarize it's the answers for the three different equations