so back to stoichiometry for just a second having covered concentration and solutions we're gonna revisit stoichiometric and the stoichiometry map so you'll remember the center of the stoic chemistry map is moles with the mole ratio there in the middle and there's a bunch of different ways we can get to that central feature we can go through mass we can go through a volume of a gas as standard temperature and pressure and we can go through representative particles but we have also learned right that molarity is equal to moles per liter so if you know the concentration and volume of a solution you can figure out how many moles of solute which is kind of the reacting feature and a solution and so if we have the volume of a solution we can get ourselves two moles through the concentration and on the other end if we know our moles of unknown and we know our volume of solution then we can figure out the concentration of a product in a solution reaction so let's take a look at how this works so how many grams of ammonium nitrate can be produced when 55.0 milliliters of a 3.2 molar ammonium carbonate solution reacts with a hundred milliliters of excess aqueous copper to nitrate so first thing you're going to need is an all stoichiometric problems is a balanced chemical equation so grab your pencil grab your paper pause the video and write an equation out and here we are with our equation volume carbonate copper nitrate both aq copper carbonate is your precipitate and then you also have a solution of ammonium nitrate and in this question we want to know how many grams of ammonium nitrate are in that solution so we start with a known and an unknown we know we have 55 milliliters of a 3.2 molar solution by the way we also know we have a hundred milliliters of aqueous copper nitrate but we know that that is in excess at least the amount of copper nitrate dissolved so we don't have enough information to get two moles of that reactant and we don't know how many grams of product we're gonna produce okay so check the map what are you starting with well you're starting with a volume of a solution and a known concentration which can get us to moles on the mole map and we're trying to find grams so let's go ahead and set this up as we set some of our problems up we have our balanced chemical equation here again we're gonna get our given pieces of information written under the appropriate reactants we are going to find our known and our unknown we're going to identify any molar masses that we need so many times we go from moles to grams so in this case it'll be ammonium nitrate so you're going to want to get that molar mass and then we're going to check the map and we're gonna see that we are gonna do a conversion here from concentration two moles to moles to grams so you can start with your concentration sorry your volume of your solution and then we're gonna use the concentration so there's our volume in the top left corner we're gonna then use the concentration to get us to moles of reactant or moles of known so the concentration is 3.2 moles of ammonium carbonate for every one liter of solution and so you can see that when we set it up this way the leaders are going to cancel out there next step is going to be from here a pretty standard stoichiometric problem we're gonna do mole ratio and the mole ratio here is one to two we're going to need that ammonium carbonate on the bottom and the ammonium nitrate on the top it's going to cancel out moles and then our last step is to go from moles to Graham we're going to use that molar mass that we calculated at eighty point zero six grams of ammonium nitrate per mole right now the one mole on the bottom and they're the moles are going to cancel and the units we have left are grams of ammonium nitrate and of course the problem starts with how many grams of ammonium nitrate can be produced multiply across the top divide across the bottom make sure you're checking your sig diggs and units in your final answer and if you go ahead and do that you should end up with 28 grams of ammonium nitrate no bonus question what's the concentration of ammonium nitrate in this final solution go ahead and give that some thought see if you can come up with a calculation and see if you can come up with a prediction pause the video while you're doing that okay so how would we go about sorting that out well we know how many grams of ammonium nitrate we have and we can use the molar mass that we had calculated to figure out how many moles that is by the way if you stop the the calculation after the second step you don't do the 80 point zero six grams you can just do a quick calculation Oh point zero five five zero times three point two times two will give you the number of moles of ammonium nitrate and then the volume of the final solution well we mixed 55 milliliters with a hundred milliliters while those volumes won't always be exactly additive they're gonna be pretty darn close particularly for low concentration solutions and as long as there's not a tremendous amount of heat absorbed or given off by the reaction so we add those together we'd get 0.15 5-0 liters of solution your number of moles point three five to notice I took more sig digs that I needed there on the number of moles but then we're gonna divide those out and we should get two point three molar solution of ammonium nitrate there will also be some copper carb carbonate precipitate in that reaction container and there should be a little bit of copper nitrate left over because it's in excess so there's a bunch of different things going on in that solution about with respect to the ammonium nitrate you should be at roughly two point three molar all right so here's another one for you to try this is an identifying a limiting reactant problem and so go ahead and get yourself a balanced chemical equation let's see if you can work through which one of these is going to be the limiting reagent come up with a with a prediction with an answer there and then pause the video while you work that out alright so let's go ahead and work through this together if you've taken some time to do it we need a balanced chemical equation first so there we have two moles of silver nitrate one mole of magnesium chloride kind of produce two moles of silver chloride and one mole of magnesium nitrate the silver chloride is going to be the precipitate in this double replacement reaction we're going to put our given pieces of information on our chemical equation there and then since we're doing stoichiometry we need to get to moles so we have two different concentrations here so we want to convert both of those concentrations to moles so for the first one there we'll just set up that molarity equation solve for N and in solving friend we should get somewhere in the neighborhood of 0.15 moles of silver nitrate on the other reactant set up the same molarity equation solve for n and you're at point eight zero eight seven five moles of silver chloride so now that we have moles of both reactants now we're just a straight stoichiometric problem then you remember when you're only asked to find the limiting reactant you can go from one reactant to the other it doesn't matter which one you start with so I'm gonna go ahead and set up my game board and I'm gonna start with the point one five zero moles of silver nitrate my mole ratio there is one two - I need the silver nitrate on the bottom moles of silver nitrate things is gonna cancel out then you get multiply across the top two by it across the bottom and when you do that you get point zero seven five zero moles of magnesium chloride and so you have to ask yourself the question do I have enough magnesium chloride and it turns out you have more than enough you only need point zero seven five you have point zero eight seven five so that means magnesium chloride is in excess and silver nitrate is our limiting reagent if we had started with the magnesium chloride we would have expected that number to come out less sorry more than 0.15 zero which would tell us that we won't have enough silver nitrate so it doesn't matter which reactant you start with you're going to end up identifying silver nitrate as the limiting reagent in either situation and so there is a quick overview of using solutions in stoichiometric problems and so now you have a new skill