[Music] good day everyone and welcome to our next lesson this is reaction stoichiometry so we will just be continuing our concepts from composition stoichiometry but now we will be applying our learnings for chemical reactions so let's proceed let's start with the definition of chemical equations because reaction psychometry will heavily rely on chemical equations these are representations of chemical reactions and they allow us to describe a reaction qualitatively qualitatively meaning we know the composition of the products and the reactants this is an example of a chemical equation 2h2 plus o2 yields 2 h2o the important parts of a chemical equation are the reactants in this case that's hydrogen and oxygen the product in this case we only have one product that is water and the coefficients that make the reaction balance for this equation we have two for hydrogen one for oxygen and two for water chemical equations should first be balanced before they can be used for stoichiometry so i'm sure that you have already encountered balancing chemical equations before we will just be doing a quick review here let's try to balance this reactions for the first one we have methane ch4 plus oxygen o2 yields co2 that's carbon dioxide plus water or h2o in balancing these kinds of reactions all you have to do is to count the number of each elements on the reactant side and on the product side as a review this is our reactant side and this are our reactants and this is our product side and these are our products so if this is your first time balancing chemical reactions you might find it easier if you draw a line between the reactants and the products and list all of the elements that you see in the reaction in this case we only have three elements that is carbon hydrogen and oxygen once you have listed all of the present elements in the chemical equation simply count how many elements of each kind do you have in the reactant and in the product side so let's start with carbon so for carbon you only have one in the reactant side that comes from the ch4 for the product side you also have one coming from the co2 so that is balanced we say that a chemical equation is balanced if you have the same number of elements in the reactant and in the product sides let's proceed with hydrogen for the reactant side you have four atoms of hydrogen coming from ch4 in the product side you only have two now the question is how can you make the number of hydrogen atoms balanced in the reactant and in the product side you can do it in two ways the first way is you can lessen the number of hydrogen atoms in the reactant side how do we do that you can multiply a fraction to ch4 so if i multiply one half to ch4 that means that the number of hydrogen would be four times one half that's equal to two and that balances my hydrogen however it would be easier to go with your second option which is to multiply the water on the product side to increase your number of hydrogen atoms remember for balancing equations you cannot add or subtract individual atoms to the chemical equations you can only multiply coefficients you can multiply whole number coefficients if you want to increase certain counts of atoms or you can multiply fractions if you want to decrease certain types of atoms so in this case let's try multiplying water by two so by multiplying water by two that means that our hydrogen in the product side now becomes four on the reactant side it's not affected so therefore we have balanced hydrogen last one oxygen if you count oxygen on the reactant side that is two and on the product side you have two from co2 and you also have two from water remember you need to multiply the coefficients with every element so you need to multiply two to both hydrogen and oxygen in water so therefore oxygen in your product side is four now our oxygen is not yet balanced we need to multiply the oxygen in the reactant side by two to also make it four and that's as simple as multiplying o2 by 2 you make the oxygen in the reactant side for if all the counts of all the elements are balanced then your reaction is balanced so we now say that the balanced equation is ch4 plus 202 yields co2 plus 2 h2o that is now our balanced equation okay so if you want to get proficient in this balancing then you need to practice and when you are already well practiced you don't even need to write this table you can perform the balancing just by looking at the equation let's do that with the next examples our next example kclo3 yields kcl plus o2 okay so how do we balance this you first recognize what are your elements in the chemical equation so again we have three elements that's potassium or k you have chlorine cl and you also have oxygen let's start with potassium so if you count the number of potassium in the reactant and in the product side we find that they both have one so potassium is already balanced it's one in the reactants one in the product let's proceed with chlorine there's also one chlorine in the reactants and one chlorine in the product so we say that chlorine is also balanced so that leaves us with oxygen we have three oxygen atoms in the reactant and only two oxygen atoms in the product so how do you balance this equation well you have you have several options one of your options is you can introduce fractions into this equation such that we will only multiply oxygen by a single fraction and the entire equation will already be balanced so the question is what can you multiply to 2 to make it equal to 3 so mathematically we are saying that if i let the coefficient of oxygen to become x so what we're really trying to do is what can you multiply to 2 that becomes 2x such that the answer would be equal to 3 y 3 because you have three oxygen atoms in the reactant side and this is this is a mathematical equation you can solve for x and we get three halves let's check if i multiply oxygen by three halves will the equation be balanced so we did not change the amount of potassium and chlorine so there's still both one in the reactant and one in the product for the oxygen in the reactant you have three and for the oxygen in the product you have three halves times two which is equal to three so this is a balanced equation there is another route though if you don't want to use fractions you can multiply whole numbers to the other components of the chemical equation so let's try to do that let's start with the unbalanced equation again let's see if we arrive with the same answer using a different method again potassium and chlorine are both balanced oxygen is not yet balanced what if i multiply kclo3 by 2 the purpose being if you multiply 2 by 3 oxygen you get 6 and 6 is an even number and an even number of oxygen atoms makes it easier for us to balance it later okay if i multiply kclo3 by 2 that means that i also need to multiply kcl by 2 to make the counts of potassium and chlorine balanced okay now for oxygen on your reactant side you have six atoms of oxygen on the product side you only have two atoms of oxygen so what do you multiply to two to make it equal to six we simply multiply two by 3 and that makes oxygen balance as well now is this chemical equation the same as our previous equation they're actually the same if i multiply the entire second equation by one half what do we get we get kclo3 yields kcl plus three halves o2 and that's the same as our first equation so you see there are multiple methods on obtaining the same balanced equation okay last one we have p4 plus cl2 yields pcl5 we only have two elements in this case that's phosphorus and fluorine let's start first with phosphorus so on the reactant side you have four atoms of phosphorus on the product side you only have one atom of phosphorus so how do we balance we want to make the product side equal to four so we simply multiply this by four however if i multiply the pcl5 on the product side with four i not only change the number of phosphorus but i also change the number of chlorine atoms so on the product side we now have 20 chlorine atoms that is four times five okay so you have 20 chlorine atoms on the product side on the reactant side you only have two so what can you multiply to 2 to make it equal to 20 you simply multiply it by 10. so let's check phosphorus you have 4 in the reactants 4 in the products for chlorine you have 20 in the reactants you also have 20 in the product so this equation is balanced okay if you cannot keep up with this type of balancing then you can do the old-school method of listing every element in the reactant and in the product side okay you just need more practice with this let's proceed now that we know how to balance equations we now need to interpret what are those numbers in the balance equation trying to tell us now remember that the purpose of the balance equation is to provide us with quantitative relationships between the components of a chemical reaction let's take this example we have the balanced equation two sodium plus chlorine yields two sodium chloride so to make sure that it's balanced you simply count you have two atoms of sodium in the reactant and two atoms of sodium in the product for chlorine as well you have two in the reactant and two in the product how do we derive relationships coming from this balanced equation you simply have to list equivalencies or we have to list ratios so let me show you how again our balance equation that's 2n a plus cl2 yields 2n a cl so what does this mean this means that 2 moles of sodium is equivalent to one mole of chlorine is also equivalent to two moles of sodium chloride so if you are running a chemical reaction that means that if i get two moles of sodium and i combine that with one mole of chlorine gas i can get two moles of sodium chloride okay now this gives us more use if we are writing it as a ratio because in stoichiometry it's all about using ratios so we can write ratio such that two moles of sodium is to one mole of cl2 that's one ratio we can use or we can also use two moles of sodium equivalent to two moles of sodium chloride that's also valid or you can use one mole of cl2 is to two moles of sodium chloride all of this are equivalent ratios which you can use in reaction stoichiometry okay you can even use them inverted so we can have one mole of cl2 is to two moles of sodium depending on the need okay how do we vary this ratio depending on the need let me give you an example let's say for example that i give you the number of moles of sodium so let's say for example this is given this is x number of moles of sodium and i want you to convert that to the number of moles of sodium chloride now this is only possible if we are talking of a chemical reaction because the substance is changing so you have sodium you want to change that to sodium chloride okay again i repeat this is only possible in a chemical reaction so if this is the case if x number of moles of sodium is you're given how do we convert that to the number of moles of sodium chloride you simply choose the correct ratio coming from your balanced equation so we will be multiplying this amount with a ratio so you have already learned this in composition stoichiometry you take the unit of the numerator in this case that's moles of sodium and you place that in the denominator of the factor so in the denominator we need to have the number of moles of sodium and in the numerator you place the unit that you want to achieve so we want to achieve the number of moles of sodium chloride in this case now take a look at the balance equation and simply fill in the numbers that you see so from the balance equation you have two moles of sodium this is that two moles of sodium that is equivalent to two moles of sodium chloride so your numerator is also two moles of sodium chloride and solving that gives you also x number of moles of sodium chloride so that is how we utilize balanced equations let's try an example let's solve the first one the thermite reaction is depicted with this reaction fe2o3 plus aluminum yields aluminum oxide plus fe so let me first write the given the question is if 25 grams of fe2o3 was mixed with a large amount of aluminum how much fe would be produced so this is our equation it was stated in the problem that you have 25 grams of fe2o3 you have a large amount of aluminum meaning aluminum is present in excess which we will tackle later we want to find out how much iron can you produce first take a look at your chemical equation and make sure that it's balanced if it is not yet balanced then you balance the equation so let's see let's count iron iron in the reactant side you have two iron in the product side you have one so we need to multiply the iron in the product side by two to make it a balanced equation next we have aluminum how many aluminum is in your reactant side it's one and how many aluminum is in your product side that's two it's not yet balanced so you multiply aluminum on the reactant side by two to make it balanced last one is oxygen you have three oxygen in the reactants you also have three oxygen atoms in the product so oxygen is balanced therefore we now have a balanced equation now you can proceed with the stoichiometry okay one thing that you need to remember in stoichiometry is that you cannot proceed with the reaction stoichiometry with grams as you're given you first need to convert the grams to moles before you can proceed with reaction stoichiometry the conversion of grams to moles was already tackled in our previous lesson so if you're still confused with that you can go back and watch that lesson so for this problem here's our plan we first get the given which is 25 grams of fe 203 again as i've said you cannot use grams in reaction stoichiometry so i need to convert this first into the number of moles of fe2o3 now what do we use when we are converting from grams to moles and vice versa we use the molecular weight of the compound in this case we are going to need the molecular weight of fe2o3 next the problem requires us to convert the amount of f8 fa203 to the amount of fe so therefore from the number of poles of fe to o3 we can now convert this to the number of moles of fe using the balanced equation and by using the balance equation we are practicing reaction stoichiometry okay so we have a two-step process but we are just going to do that in one straight calculation if you have a two-step process like this that means that you have two factors that you need to figure out the first factor or the first conversion factor is the molecular weight of fe2o3 the second factor is the coefficients coming from the balanced equation let's first determine the molecular weight of fe2o3 here we have again our periodic table so let's first list the molecular weights of iron and oxygen then we can compute for the molecular weight of f8o3 for iron we have this value 55.845 and for oxygen we have 15.999 you can also round off this values no it would have it would have just a little effect on the final answer but otherwise you will be very very close so you can use 16 grams per moles for the for the molecular rate of oxygen and you can use 55.85 for the molecular weight of iron okay using this data we can compute for the molecular weight of fe2o3 that is simply two times the molecular weight of iron plus three times the molecular weight of oxygen so let's compute we have 55.845 times two for iron plus for oxygen 15.999 times 3. so the molecular weight of fe2o3 is 159.687 grams per mole okay so we can now proceed with the stoichiometry so the first half of our solution is from composition stoichiometry wherein we simply convert from grams to mole the second half of our solution is reaction stoichiometry so again we start with a given so you are given 25 grams of fe2o3 we want to convert this to moles take a look at our given you're given grams in the numerator here therefore you need to have grams fe 203 in the denominator of the first factor that gives you mole of fp2o3 at the numerator of the first factor and this is the molecular weight so therefore we need to divide by the molecular weight that is divided by 159.687 grams per 1 mole okay that's the first factor you see that grams of fe 203 will cancel for the next factor we now take a look at the balanced equation so taking a look at the balance equation and then taking a look at your first factor your remaining unit is moles of fe2o3 that means that in the second factor moles of fe2o3 should be in the denominator and what we are looking for we are looking for the amount of fd let's assume that that is the number of moles of f e that should be in the denominator why did we do that such that we can sell moles of fe 203 are only left with the amount of fe now in this case the problem only asks us how much iron is produced from the reaction it did not specify if it was asking for grams of iron or moles of iron so in this case we can just stop at moles of iron well anyway if you want to get grams of iron you simply have to convert from moles of iron to grams of iron okay now what is our ratio from the balanced equation from the balance equation f e to o three that's one and for ion that is two we have determined those coefficients from balancing the equation okay so you see the balancing of the equation is very important if you did not balance your equation prior to solving then you are automatically wrong now you can solve this equation so we start with 25 grams we divide by the molecular weight 159.687 and you multiply by 2. this is the number of moles of iron produced 0.313 moles of fe we retain three significant figures because our given 25.0 grams is three significant figures okay so that is how we apply a combination of composition and reaction stoichiometry to problems let's take a look at the next example in the combustion of propane that's c3h8 it reacts with oxygen gas and produces carbon dioxide and water how much water is produced from the combustion of two kilograms of c3h8 now in this case you are not given the balanced chemical equation you need to figure out the chemical equation from the worded problem so it said that propane or c3h8 reacts with oxygen and produces co2 and h2o those are your keywords if the problem says that it reacts that means that those are part of the reactants and if the problem says that it produces then those are part of the product so let's write the given it says that c3h8 or propane reacts with oxygen to produce carbon dioxide that's co2 and water or h2o so this is our chemical equation now we just have to balance it how do we balance let's first determine all of our atoms present we have three that is carbon hydrogen and oxygen let's first balance carbon in the reactant side you have three atoms of carbon that is c3 okay in the product side it's only one coming from co2 so how do we balance that we multiply co2 by 3 such that our carbon is now 3 on both sides so carbon is now balanced next is hydrogen in the reactant side you have eight atoms of hydrogen coming from c3h8 and on the product side you only have two coming from h2o so how do we balance that what do you multiply to 2 to make it equal to 8 we multiply 4 such that you also have 8 hydrogen in the product side okay final one oxygen you only have two oxygen on the reactant side and how many on the product side that is three times two from co2 that's six plus four from water that is ten so we have ten atoms of oxygen on the product side what do you multiply to 2 to make it equal to 10 we multiply it by 5. so this is now our balanced equation okay do not proceed with your calculations unless your chemical equation is balanced so we are given 2 kilograms of c3h8 and what's it asking for the problem is asking us to determine how much water is produced coming from the two kilograms of c3h8 okay so we want to convert from c3h8 to water we simply follow our approach from the previous problem we're in from the kilograms of c3h8 which is our given we want to convert this to the number of moles of c3h8 using its molecular weights and then now we can use reaction stoichiometry to convert from number of moles of c3h8 to number of moles of h2o the problem is just asking us for the amount of water produced with no specifics if it's asking for moles or if it's asking for grams so in here we can stop with the number of moles of water but in our exams you will be given specific instructions on what unit should you have on your final answer so for our first step from kilograms of c3h8 to the number of moles of c3h8 what conversion factor are we going to use we simply need the molecular weight of c3h8 because we're simply converting from grams to mole or from kilograms to mole for the second step we are now converting from mole of c3h8 to mole of water so we are changing our specie this pertains to the chemical reaction therefore we need the coefficients of the balance equation for the second step okay so this is the same process as with our previous example let's first determine the molecular weight of c3h8 the molecular weight of carbon is 12.011 you multiply that by 3 plus the molecular weight of hydrogen is 1.008 we multiply that by eight the molecular weight of propane is 44.097 grams per mole we can now use this for our conversion okay let's begin the conversion now you are given two kilograms of c3h8 if you take a look at the unit for your molecular weight that is grams per mole it's not kilograms per mole so it would be wise if we simply convert kilograms to grams before proceeding and the conversion factor is easy one kilogram is equivalent to 1000 grams of the material so let's use that as our first factor for our first factor we say that one kilogram is equivalent to 1000 grams we can cancel the kilograms so we now have 1000 grams of c3h8 next factor we want to use the molecular weight to convert from grams to mole so since grams is in the numerator here grams of c3h8 should be your denominator in the first factor and the numerator should be mole of c3h8 so we need to divide by 44.097 grams per mole okay grams would cancel the last factor would make use of the balance equation to equate the number of moles of c3h8 with the number of moles of h2o so since on our last factor c3h8 is in the numerator on the next factor it should be on the denominator so that we can cancel it and the numerator should be the one that we are looking for that is the number of moles of h2 and then take a look at the balance equation to find the coefficients for c3h8 that is one and for h2o that is four moles of c3h8 will cancel and we are left with the unit that we are desiring we want to get the number of moles of water okay let's solve that that is 2 multiplied by 1000 that's for the first conversion and we divide by the molecular weight 44.097 for the second factor and for the last factor we multiply by four so our answer is 181 moles of h2o we retain three significant figures because the given is three significant figures i will be releasing a separate video to show you other examples for this topic okay let's have the last example the haber process produces ammonia by combining nitrogen and hydrogen gas at high pressures and temperatures the limiting reactant is nitrogen if 100 kilograms of ammonia is required how much nitrogen should we use so this is another example of a worded problem wherein you're not given the chemical equation and we have to decipher the chemical equation from the problem so the problem stated that you produce ammonia that means that nh3 is your product the problem also mentioned what did it use to produce your ammonia it says that you use nitrogen gas and hydrogen gas however you have to know that nitrogen and hydrogen gases are both diatomic molecules diatomic meaning that they naturally exist as two atoms merge together now just a side topic we have a list of seven diatomic elements meaning at their natural state they are diatomic so that you will not forget because in worded problems like this you need to know which elements are diatomic and which elements are not so here are the seven diatomic elements hydrogen nitrogen fluorine oxygen iodine chlorine and bromine so how do you remember this list of the seven diatomic elements we have a mnemonic pay attention have no fear on ice cold beer that is our mnemonic okay just remember have no fear on ice cold beer that is our mnemonic for the naturally occurring diatomic elements okay that's why in this problem when it mentioned that we use nitrogen gas and hydrogen gas we don't use n only we don't use h only okay that's wrong we need to use the diatomic forms n2 nh2 okay now moving on so what do we do with this first we balance the equation so we only have two atoms here that's nitrogen and hydrogen if you count the number of nitrogen from the reactant side you have two on the product side you only have one so we multiply the product by 2 such that nitrogen will be balanced next is hydrogen on the product side you have 6 hydrogen that is 2 times 3 so what can you multiply to 2 to make it equal to 6 we multiply it by 3. so this is now our balanced equation so this is a reverse calculation wherein you are given the product you are given 100 kilograms of ammonia and we want to know how much nitrogen is required let's first write that the limiting reactant here is nitrogen and then perform stoichiometry as we have learned what is our plan here so our plan is start from the given 100 kilograms of ammonia first of course we want to convert this to grams just as what we did to the previous problem we want to convert grams of ammonia to the number of moles of ammonia because again we cannot use grams in our stoichiometry now what do we use to convert from grams to moles we use the molecular weight of ammonia now that we have the number of moles of ammonia we can convert that to the number of moles of nitrogen using the balanced equation and that is our process from the mass of ammonia convert that to moles of ammonia and then from moles of ammonia convert that to moles of nitrogen gas okay let's first determine the molecular weight of ammonia the molecular weight of nitrogen is 14.007 plus hydrogen 1.008 times 3 that is 17.031 grams per mole now we can proceed with our calculation you start with a given 100 kilograms of ammonia again just as what we did from the last example we want to convert this first to grams so one kilogram is equivalent to 1000 grams of ammonia okay first factor is the molecular weight so again grams in the numerator here so in your factor grams of ammonia should be in the denominator and moles of ammonia in the numerator so effectively we are dividing by the molecular weight of 17.031 next factor we take a look at the balanced equation so moles of ammonia should be in the denominator while the one that we are looking for moles of nitrogen should be in the numerator so from the balance equation you have two moles of ammonia equivalent to one mole of nitrogen gas take a look at the cancellation of units so kilograms cancel grams cancel moles cancel calculating this is 100 times 1000 divided by 17.031 divided by two our answer we can only keep three significant figures assuming that the two zeros in one hundred are significant so our answer is two thousand nine hundred forty moles of nitrogen assuming that the zero here is not significant okay so that is how we perform stoichiometry let's proceed now let's dive into the concept of the limiting reactant since we have already introduced it not all reactions totally exhaust their reactants so consider this reaction here we have one of our previous example no that's c3h8 plus o2 yields co2 plus h2o so let's balance that first c3h8 plus o2 yields co2 plus h2o to balance we multiply co2 by 3 we multiply h2o by 4 and then oxygen we have 10 on the product side so we multiply the reactants by 5. okay let's go back if we start with one mole each of c3h8 and o2 how much co2 would be formed okay let's try to solve that first the problem says that what if we start with one mole of c3h8 and one mole of o2 how much co2 can we form okay let's perform some quick stoichiometry let's start with one mole of c3h8 now this is a straightforward conversion because we already have balanced our equation so in the denominator we need moles of c3h8 in the numerator we have moles of co2 because that is our unknown now based on the balanced equation one mole of c3h8 is equivalent to three moles of o2 therefore we can form three moles of co2 what happens if we start with one mole of o2 we need to have moles of o2 in the denominator and moles of co2 in the numerator because again that is our unknown so based on the balance equation for every five moles of o2 there are three moles of co2 that means that we have three-fifths moles of co2 if we start with one mole of co2 now take note that we have two answers this one and this one the question is which of this is the correct answer because we can only have one correct answer we have the higher answer three moles and we have the smaller answer three-fifths of a mole now the correct answer here is the smaller one the three-fifths of a mole co2 and the three moles co2 is wrong now why is that that is because of the concept of the limiting reactants that entails that one of your reactants would be the first one to be consumed and when one of your reactants is fully consumed the reaction will stop and you cannot get more than what the limiting reactant can produce let me let me discuss the concept of the limiting reactant with a simple example imagine that you're having a party and you are serving hot dog sandwiches how do you make a hot dog sandwich for a hot dog sandwich of course you need your hot dog and you need a hot dog bun and if you combine those you get a hot dog sandwich okay in terms of a chemical equation this is one hot dog plus one bun would give you one hot dog sandwich the rule is you cannot place two hot dogs on one bun you cannot do that and you also cannot place half of a hot dog to one bun because someone would get angry so it's always one hot dog plus one bun would give you one hot dog sandwich so what if you start with 20 hot dogs and you combine that with 20 buns what do we get of course we get 20 of the hot dog sandwich so that's fine because it's one is to one and 20 is the 20 when reduced becomes one is to one but what if you were able to buy 30 hot dogs but you were only able to buy 20 hot dog buns the question is how many hot dog sandwich can you produce can you produce 30 hot dog sandwiches of course not because you are missing 10 buns that means that you can only produce 20 hot dog sandwich and we say that in this case your bone is the limiting reactant because that's the first reactant to become exhausted and your hot dog in here is called the excess reactant because when you have already exhausted your hot dog buns you would still be left with 10 hot dogs in this case okay i hope this simple example makes sense to you we can also reverse this and let's say that we only have five hot dogs but we have 10 hot dog buns so how many hot dog sandwiches can we produce of course we cannot produce 10 hot dog sandwiches because you will be lacking 5 hot dogs the answer here is you can only produce five hot dog sandwiches your hot dogs in this case is the limiting reactant and the bond is now your excess reactants so that is a simple explanation of the concept of the limiting reactants now when you are solving for problems involving chemical equations you just go back to this example so that you will be reminded of the concept of the limiting reactants okay let's go back let's solve this example sodium violently reacts with water to form sodium hydroxide and hydrogen gas if 3.4 grams of sodium is reacted with 20 grams of h2o which is the limiting reactant and how much sodium hydroxide would be produced again this is a worded problem so we have to convert it to a chemical reaction sodium reacts with water that means that sodium and water are our reactants to form that is your keyword for products sodium hydroxide and hydrogen gas so we can write that as sodium reacts with water to form sodium hydroxide and hydrogen gas so you have to review your nomenclature for you to be able to convert the names of the compounds to their chemical formula okay and then the problem states that we have 3.4 grams of sodium and then 20 grams of water okay before we proceed let's first balance our equation we have three elements here sodium hydrogen and oxygen let's take a look at sodium so sodium is already balanced you have one in the product and one in the reactants next is hydrogen you have two in the reactants and you have three in the product so how do we deal with that one thing that we can do so that we will no longer affect the coefficients of the other compounds is to simply multiply your h2 with one half so what does that accomplish that makes the total number of hydrogen in your product two that is one from the sodium hydroxide and one from hydrogen that's balanced with the amount of hydrogen in the reactant side okay if you don't want this balanced equation you can multiply the entire equation by two we can have 2n a plus 2 h2o yields 2 naoh plus 1 hydrogen or 1 h2 okay you can use either of these two equations but let's use the first equation the problem is asking us which of the two reactants is the limiting reactance and how much sodium hydroxide can we produce our approach here would be to determine the amount of sodium hydroxide using both of the reactants and whichever gets the lower amount of sodium hydroxide then that is our limiting reactants okay let's start with sodium so we have 3.4 grams of sodium now remember we are given grams of sodium we cannot use grams in stoichiometry so we have to convert grams to moles using the molecular weight of sodium so let's look for the molecular weight of sodium okay the molecular weight of sodium is 22.99 so we can divide this by the molecular weight convert it to moles of sodium now that we have converted from grams to moles let's now take a look at the balanced equation from the balance equation we see that for every one mole of sodium there is one mole of sodium hydroxide so it's just one is to one okay solving what do we have 3.4 divided by 22.99 that's 0.148 mole of sodium hydroxide let's set that aside first and let's solve using water we start with 20 grams of water and again we convert this to moles the molecular weight of hydrogen is 1.008 times 2 plus for oxygen 15.999 that's 18.015 grams of water is equivalent to one mole of water and then let's visit the chemical equation for every one mole of water it's equivalent to one mole of sodium hydroxide as well cancel the units you're left with moles of sodium hydroxide solving what do we get 20 divided by 18.015 answer 1.11 moles of sodium hydroxide now that we have two values for the amount of sodium hydroxide in the product we need to choose which one is correct because we cannot keep both so again our rule is you choose the one with a lesser amount of product produced this is our amount of sodium hydroxide produced this is wrong and therefore we say that sodium is our limiting reactants and water is our excess reactants so that's how you answer these types of problems okay you perform calculations for the two reactants you choose which one produces the lesser amount of products and the lesser number is the correct amount of product and the given amount used to produce the lesser amount of product is your limiting reactance and everything else is the excess reactants okay let's proceed let's discuss the concept of percent yield so percentile is normally being used to compare the theoretical yield with the actual yield obtained from an actual experiment the main principle behind determining the percentile is that not all reactions go to completion or sometimes reactions can go to completion but we cannot recover all of the products the percent yield can be used to quantify the completeness of the reaction and this is our formula the percent yield is equal to actual yield over theoretical yield now just some notes the actual yield in this case would be the actual amount of the substance that was recovered through an experiment so the only way for you to have a value of the actual yield is that if you perform the experiments and you actually measured the amount of the products well the theoretical yield is the amount of the product that was computed from stoichiometry that's why it's theoretical because it's from our calculations okay a percent yield of 100 indicates that the actual yield is the same as the theoretical yield meaning that the reaction is complete okay let's try to solve this example determine the percent yield if 5 grams of sodium hydroxide is recovered from the previous reaction from our previous example our given equation is sodium plus h2o yields naoh plus h2 to balance the equation that's multiply h2 by one-half and we also learned that sodium is our limiting reactant and water is our excess reactants we were provided with 3.4 grams of sodium 20 grams of water although we now know that 20 grams of water is in excess for this reaction and with that we have produced 0.148 moles of sodium hydroxide we want to determine the percent yield of the reaction if 5 grams of sodium hydroxide is recovered that means that this given 5 grams of sodium hydroxide recovered is our actual yield from the reaction so let's write the actual yield is 5 grams of sodium hydroxide and this amounts the 0.148 moles of sodium hydroxide that we have obtained based from our calculations is our theoretical yield now before we can compute for the actual yield we first have to convert the theoretical yield to grams or you can convert the actual yield to moles okay you just need to have the same units for the actual yield and the theoretical yield so for this one let's convert the actual yield to moles so we have five grams of sodium hydroxide let's convert it to moles first let's determine the molecular weight of sodium hydroxide so sodium is 22.99 oxygen is 15.999 hydrogen is 1.008 therefore the molecular weight of sodium hydroxide is 39.997 grams of sodium hydroxide that's equivalent to 1 mole of sodium hydroxide that's 5 divided by 39.997 we have 0.125 moles of sodium hydroxide remember that this value is still our actual yield because we simply converted the actual yield of 5 grams into moles okay now we are ready to solve for the percent yield for the percent yield you simply get the actual yield divide by the theoretical yield times 100 so that is 0.125 moles for the actual yield divided by 0.148 moles for the theoretical yield times 100 percent we can now solve for the percent yield 0.125 divided by 0.148 times 100 that is 84.5 percent so we can say that 84.5 of the product was recovered from the reaction always remember for problems involving percent yield the actual yield cannot be higher than the theoretical yield okay if that happens please reject your calculations because you might be wrong on the determination of theoretical yield if the actual yield is higher than the theoretical yield okay the highest possible value of actual yield is the theoretical yield such that your percent yield is 100 for the last part of this lecture we need to talk about the types of reactions so this is just some extra content no you just have to be familiar with the most common types of reactions so i have listed the four most common that is combination decomposition single displacement or double replacement let me give you some examples for combination reactions or addition reactions the general form of combination reactions is a plus b yields c where in a and b could both be either elements or compounds and c is a compound okay combination reactions are very easy to spot because you have two or more reactants and you only have one product an example of that is the formation of water so for the formation of water that's hydrogen gas plus oxygen gas yields h2o balance the equation multiply o2 by one half so this is a type of a combination reaction okay another example is the formation of sodium chloride or common table salt that is sodium plus chlorine gas yields sodium chloride to balance the equation multiply cl2 by one half okay next type the composition the composition reactions are the opposite of addition reactions so if addition reactions have many reactants and only one product the composition reactions have one reactant in many products the general form is a decomposes to b plus c okay you can have more than two products so in this case your reactants or a is a compound while your products b and c could both be either compounds or elements okay an example of a decomposition equation is the decomposition of calcium carbonate upon introduction of heat so we have calcium carbonate caco3 that decomposes to calcium oxide and carbon dioxide so you see here that our products are both compounds no and our reactant is also a compound and we signify the addition of heat into the reaction by the addition of a small triangle on top of the yield sign okay another example is the electrolytic decomposition of sodium chloride into its constituent elements so sodium chloride if it's molten and if it is electrolyzed will form into sodium metal and chlorine gas multiply one half to chlorine to balance the equation okay and you will notice that our second example here is simply the reverse of the addition reaction earlier okay so the composition and addition reactions are simply reverses of each other next equation single displacement for single displacement reactions we have the general form of a plus b c yields a b plus c so what's happening here is a is replacing c as the partner of v so it's like one is displacing the other in the compound okay so a could be an element it could also be an ion and bc needs to be a compound so let me give you an example of a single displacement reaction the reaction between hcl which is an acid with a metal such as magnesium is a single displacement reaction because magnesium displaces hydrogen such that magnesium will now be partners with chlorine so our products here would be magnesium chloride plus hydrogen gas now to balance the equation we simply multiply this by two okay so it's called single displacement because only one specie is replacing one of the ions in the compound and single displacement reactions normally happen in aq solutions okay for single displacement reactions we are following what we call the activity series of metals so let me show you the activity series of metals okay this is an example of an activity series so the way on how we read this activity series is that metals or elements at the bottom are less reactive and metals at the top are more reactive so something at the top let's say for example magnesium in here magnesium can replace anything that is placed below it so magnesium can replace aluminum magnesium can replace iron it can replace hydrogen that's why in our example here magnesium was able to replace hydrogen because magnesium is higher in the activity series than hydrogen okay if you take the reverse and let's say for example you want copper to replace hydrogen there will be no reaction because hydrogen is higher in the activity series than copper so how do we write that let's say for example you have the reaction hcl plus copper and you want copper to replace hydrogen in this case so there will be no reaction because again because copper is lower in the activity series than hydrogen okay so there's no need to memorize this this activity series whenever we'll be having an activity or in the exam you will be given a copy of the activity series table you just need to familiarize yourself with this types of reactions okay last one the double replacements reactions so for double replacement reactions our general form is a b plus c d yields a d plus c b so it's just like a change in partners the partner of a is originally b and then after the reaction the partner of a is now d an example of a double replacement reaction is this so we have calcium chloride let's say for example we add that to sodium sulfate and our product would be calcium sulfates and sodium chloride to balance this equation we place two for sodium chloride you might be wondering how did i come up with these types of products you have to be familiar with the cat ions and ions for you to be able to determine what are the products of such reactions let's say for example for calcium chloride i know that the cat ion is calcium 2 plus because calcium is a group 2 metal so this should be part of your previous learnings in chemistry and the chloride ion is cl minus okay if i crisscross calcium and chloride what we would get is the two would be the subscript for chlorine and then the one would be the subscript of calcium so we get ca1 cl2 hence the calcium chloride okay but in this case we want the halogen chloride to take on different partners so for sodium sulfate the cat ion is sodium plus one because sodium is a group one metal and the entire anion sulfate is so4 with a charge of two minus so the cat ion on the first compound that's calcium will be joined by the an ion of the second compound that is sulfate so if we combine calcium two plus and then so4 two minus if we perform the crisscross the two will be the coefficient of sulfates and two will also be the coefficient of calcium we have ca2 so4 quantity 2 but since we can simplify this this becomes ca so4 hence caso4 is one of our products okay the same can be said through for nacl so you have n a here and c l here if you combine them then you form n a c l so that is how double replacement reactions work you have to review your previous learnings in chemistry this marks the so this marks the end of our lesson in reaction stoichiometry however we do have several more videos coming and those will be example videos and we also have two videos coming detailing solution stoichiometry and stoichiometry involving gases so watch out for those also please check your blackboard accounts for your activity this week okay that's the end of this lesson i hope you have learned something please practice solving problems and as always keep safe [Music] you