in this video we're going to go over limiting reactants excess reactants how to find out how much excess reactant is left over how to calculate the thetical yield and the percent yield so let's begin two moles of propane reacts with eight moles of oxygen gas in a combustion reaction how many moles of carbon dioxide are formed so what's the first thing that we need to do here the first thing that you should do is write a balanced reaction action propane is C3 h8 oxygen gas is O2 it's diatomic carbon dioxide is CO2 and for any combustion reaction water is always a product now before we can do any sort of calculations we need to balance this equation whenever you're balancing a combustion reaction balance the carbon atoms first and then the hydrogens and save the oxygen atoms for last so notice that we have three carbon atoms on the left side so we need to put a three in front of CO2 notice that we have eight hydrogen atoms on the left but two on the right 8 / 2 is four so we need to put a four in front of H2O now how many oxygen atoms do we have in total on the right side 3 * 2 is six so we have six oxygen atoms from the three CO2 molecules and the subscript here is one if you don't see a number 4 * 1 is 4 so we have four oxygen atoms from the four water molecules so we have a total of 10 10 ID by this two is five so we need to put a five in front of o2 and so the reaction is now balanced so we have 2 mol of propane and 8 mol of oxygen which one is the limited reactant is it propane or oxygen The Limited reactant is the one that runs out first to find it if you have the moles of both reactants one method in which you can use is simply divide the moles by their respective coefficient it's not going to be the always the one with the lower moles it's the one with the lowest mole per coefficient ratio so for propane divid 1 for oxygen ID 5 2 ID 1 is 2 8 ID 5 is 1.6 therefore O2 is the limiting reactant propane is the excess reactant another way in which you can identify the elimin in and the excess reactant is to calculate the theoretical yield or the amount of product that can be formed in a reaction the theoretical yield is the maximum the the absolute maximum amount you can get in a reaction so to answer the first part of the question we need to find how many moles of carbon dioxide are form that's basically asking for the theoretical yield which can be in moles or grams but for this problem we're going to do everything in moles so one way you can identify the limit reactant is you can take the moles of each reactant and and calculate the thetical yeld the one that gives you a lower value that's the limited reactor the one that gives you the higher value is the excess reactant so let's start with 2 moles of propane and let's convert it to moles of CO2 whenever you want to convert from one substance to another us in a balanced reaction you need to find the ratio the m ratio between propane and CO2 is 1 to 3 now notice that we have C3 h8 on the top so we got to put C3 h8 on the bottom and CO2 on top so according to the m ratio for every one Mo of propane that reacts three moles of CO2 are produced so these units cancel and we're left with moles of CO2 since we have a two and three on top we need to multiply two and three if you have a number on the bottom you got to divide 2 * 3 is 6 so what this means is that if all of the two moles of propane reacts six moles of CO2 can be formed that's the maximum amount of carbon dioxide that can be produced from two moles of propane so now let's see how many moles of propane we can get if all of the eight M of oxygen gas reacts let's start with 8 mol of o2 over1 and let's use the M ratio to convert to CO2 so the M ratio this time it's 5 to 3 so for every five moles of o2 that reacts three moles of CO2 are produced so these units cancel 8 / 5 is 1.6 1.6 * 3 is 4.8 16 * 3 is 48 so 1.6 * 3 is 4.8 so if all of the oxygen molecules react we can get at most 4.8 moles of CO2 so in this reaction when we mix propane and O2 the amount of CO2 that we're going to get is the smaller of these two numbers so the actual threal yield is 4.8 moles of CO2 so that's the answer to the first question now the reactant that gave us this answer which is O2 that is the limied reactant the excess reactant will give you a larger uh theoretical yield which is not the correct answer so O2 is the limited reactant propane is the excess reactant and if radical yield of CO2 in moles is 4.8 now let's say if we did the experiment and we only measure 4.5 moles of CO2 what is the percent yield of this reaction the percent yield is equal to the actual amount ID the threal yield time 100% so the actual amount or the actual yield the act the amount that we actually got the experiment is 4.5 moles the maximum amount or the theoretical yield that we can possibly achieve is 4.8 that's the most that we can get so if you multiply that ratio by 100% you're going to get the percent yield so if you type it in you should get about 93.75% that's the percent yield for the reaction now if that's the percent yield what is the percent error to find it it's going to be 100us 93.75 so the percent error is 6.25% now how can we answer the second part of the question how much of the excess reactant is left over how would you figure it out feel free to the try here's the main idea of what you want to do the amount of excess reactant that is left over after the reaction is going to be the total amount of excess reactant that we had to begin with that's the two moles of propane minus the amount that actually reacts or the amount that actually participates in the reaction and that's going to equal the amount that's left over so for example let's say if we have 10 moles of excess reactant but only seven moles react therefore three is left over 10 - 7 is 3 and so that's the main idea of what we need to do to find out how much is left over so we already have the total in this case the total is 2 moles of propane we need to find out how much reacts and we can do that using stochiometry so let's see how much reacts with the eight moles of oxygen so what you want to do is start with the limit reactant and convert it to the excess reactant so the limit reactant is the 8 moles of o2 and let's convert it to the moles of propane the mol ratio between propane and O2 it's 1 to five so for every five moles of o2 that reacts one mole of propane reacts with it so 8 / 5 is 1.6 so this number represents the amount of propane that actually reacts or participates in a reaction that's how much propane is consumed in a reaction so the amount that's left over is the total minus the amount that reacts so 2 minus 1.6 is4 so that's how you can calculate the amount of excess reactant that is left over let's try this one 50 gr of benzene is placed in a container with 160 G of oxygen gas after the reaction 30 G of water were collected what is the percent yield how much excess reactant was left over after the reaction so the first thing that we need to do is write a Balan reaction so we have another combustion reaction so Benzene plus oxygen gas is going to produce water and carbon dioxide now whenever you want to balance a combustion reaction remember balance the carbon atoms first then the hydrogen atoms and save the oxygen atoms for last so we have six carbon atoms on the left side what number do we need to put in front of CO2 so we need a six and we have six hydrogens on the left side so 6 / 2 is three now notice that we have an odd number of oxygen atoms on the right side we have three from the 3 H2O molecules and 12 from the 6 CO2 molecules since 6 * 2 is 12 so if we balance it now 15 / 2 is a fraction it's going to be 15 over 2 so to void a fraction let's double everything so instead of this being a one let's make it two so we have 12 carbons on the left side so we now need a 12 in front of CO2 we also have 12 hydrogens so we need a six in front of H2O so at this point we have a total of 30 oxygen atoms 30 / 2 is 15 and so we could put the 15 in front of o2 so now the reaction is balanced now the first thing that we need to do is find the percent yield so we need the actual yield and the thetical yield we can calculate the thetical yield but typically the actual yield is usually given in a problem now there's three numbers that we have the 50 the 160 and the 30 which one is the actual y the actual yield is associated with the product Benzene is a reactant O2 is a reactant so water is a product so the 30 G of water represents the actual yield so we'll use that later but let's find the theoretical yield of the same substance water now since we have the information for both reactants we don't know which which one is limited or which one is excess there was two ways we in which we could figure this out we're going to use the second method in which we're going to calculate the grams of water that can be formed from each reactant and whichever gives us the lower number that's going to be the thetical yield and the reactant that gave us that number is the eliminate reactant so let's start with Benzene let's see how many grams of water can be produced if all of the 50 g of benzene reacts so let's start with 50 g of C6 H6 so what is the mol mass of C6 H6 so we have six carbon atoms times the atomic mass of 12 plus 6 hydrogen atoms times the atomic mass of 1 6 * 12 is 72 + 6 that's 78 so there's 78 G of benzene for every Mo of benzene so now our next step is to convert moles of benzene to moles of water using the M ratio so the ratio is 2 to six so for every six moles of water that's produced in the reaction 2 moles of benzene reacts so G of benzene cancels and moles of benzene cancel so now let's convert moles of water into GR of water so what's the M mass of water so we have two hydrogen atoms plus an oxygen atom 12 I mean 2 plus 16 is 18 so the M mass is 18 G of water per one mole so now let's do the math 50 / 78 * 6 / 2 * 18 you should get 34.6 grams so I'm going to put this number over here so if all of the 50 gram of benzene reacts 34.6 gr of water will be produced let's do the same thing for oxygen let's see how many grams of water can be produced if all of the 160 G of oxygen reacts so let's start with 160 G of o2 now the molar mass for O2 is about 32 it's 16 * 2 so there's 32 G of o2 per 1 mole of o2 now at this point we need to use the M ratio to convert from O2 to water so it's 15 to six so for every 15 18 moles of o2 that reacts 6 moles of water are produced the last part is going to be the same the molar mass is still 18 G so just like before grams of o2 cancel moles of o2 will cancel as well and moles of water so it's going to be 160 / 32 * 6 / 15 * 18 so you should get 36 G of water so notice that Benzene produces a lower amount of product so 34.6 is the thetical yield which means Benzene is the limiting reactant and and O2 is the excess reactant now that we have the thetical yield which is 34.6 G of water we can calculate the percent yield the percent yield is the actual yield of 30 G which is that number divided by the theoretical yield of 34.6 G times 100% so you should get a percent yield of 86.7% so now we could focus on the second part of the question and that's calculating the amount of excess reactant that's left over so remember to find the amount of excess reactant left over we need to take the total which is the 160 G of o2 since O2 is the excess reactant and we need to subtract it by the amount that reacts and that's going to equal the amount that is left over so let's find out how much O2 actually reacts or is consumed in a reaction what we need to do is start with the grams of the limitting reactant and convert it to the gram of the X reactant that's going to tell us how much of the excess reactant reacts with all of the limied reactant so let's start with 50 g of benzene C6 H6 and let's convert it to the grams of o2 so the M mass of benzene is still 78 G per one Mo so now at this point we need to use the mo ratio between the limit reactant and the excess reactant it's 2 to 15 so for every two moles of benzine that reacts 15 moles of o2 reacts with it and now we need to convert moles of o2 back to G and we know that the M mass is 16 * 2 or 32 so 50 / 78 * 15 / 2 * 32 is about 153.07291 so 6.2 G of the excess reactant is left over so that is it for this video thanks for watching and have a great day