we're told methanol which is used as a fuel in racing cars and fuel cells can be made by the reaction of carbon monoxide and hydrogen so this is the methanol right there they're giving us 356 grams of carbon monoxide so carbon monoxide we have 356 grams of it and they're giving us 65 grams of hydrogen of molecular hydrogen 65 grams they're mixed and allowed to react and they say what mass of methanol can be produced and then they say what mass of the excess reactant remains after the limiting reactant has been consumed so that tells you this is a limiting reactant problem that we have too much or too little of one of these two reactants these are the two reactants there the one that we have less of is the limiting reactant that'll dictate how much of the product we can produce and the one that we have more of is the excess reactant but first we have to figure out which is the limiting and which is the excess and before we even do that we should always check that our equation is actually balanced so let's just check that on the left hand side of this equation we have one carbon right there on the left hand on the right hand side we also have one carbon so that looks good so far on the left hand side of the equation we have one oxygen and on the right hand side we have one oxygen looks good so far left hand side we have four hydrogens two times two on the right hand side we have four hydrogens three plus one so this is balanced it is balanced so we can proceed to try to figure out what the limiting reactant is so the way we wanna do it is figure out how many moles of each of these were given and then figure out how many moles the stoichiometric ratio that's required by this reaction we already know what it is we know that for every two moles two moles of hydrogen required we require we can see it right here from the equation one mole of carbon monoxide one mole of carbon monoxide required this is what the equation tells us let's see how much how many moles of hydrogen and how many moles of carbon monoxide we have the first place to start and we've done this several times is what is the molecular weight of carbon monoxide so carbon monoxide molecular weight let me write here molecular weight for carbon monoxide it's a carbon has a molecular weight of 12. it's good to memorize that carbons and oxygens and hydrogens show up so frequently and then oxygen is 16. oxygen is 16 so it's equal to 28 molecular weight of 28 which tells us that carbon monoxide if we if we want to view this way so let's do the carbon monoxide first so we're dealing with the carbon monoxide so we have 356 grams of carbon monoxide and we want to write this in terms of how many moles of carbon monoxide do we have so what we want is how many how many grams per mole we'll actually write that i guess one mole has how many grams so one mole one mole of carbon monoxide has how many grams well it's molecular is 28 so if we have a mole of them if we have that you know six point oh two to twenty ten to the 23rd uh uh molecules of carbon monoxide that's going to weigh 28 grams so one mole of carbon monoxide is 28 grams or or i guess yeah one mole for every 28 grams the reason why i wanted to put the grams in the denominator so it cancels out over here so those cancel out when we multiply and so we are left with 356 times one divided by 28 is 12.7 moles of carbon monoxide now let's do the same thing for the hydrogen let's do the same thing for the hydrogen we have 65 grams of molecular hydrogen what's the molecular weight of hydrogen i'll do this in green over here molecular weight of hydrogen is well each hydrogen atom has a liquid of one times two which is equal to two so we have 65 grams of molecular hydrogen and the same way we want to write it in moles so we're going to say we're going to multiply it times the one mole of hydrogen is equal to how many grams well we just figured out one mole is equal to its molecular weight is two so a mole of it is going to have a mass of 2 grams so you could view this as 2 grams per mole or 1 mole per 2 grams and we want the grams in the denominator so it cancels out over here and so let's do the math that cancels out with that and we have 65 times 1 divided by 2 65 divided by 2 is what 32.5 32.5 32.5 moles of hydrogen of hydrogen moles of hydrogen now we know exactly how many moles of carbon monoxide and how many moles of hydrogen they've given us let's figure out what the ratio is and we'll do it in the exact same way as we wrote up here so we have 32.7 32.5 this is what we're given moles of hydrogen let me write that a little bit neater moles of hydrogen and we're given 12.7 moles i'll do the same color 12.7 moles of carbon monoxide so if we were to just divide this what is this i guess you could imagine divide the numerator and denominator by 12.7 so this can be rewritten as so if i just rewrite this i should have written it here to begin with i could write this as we have 2.56 moles of of molecular hydrogen for for every one mole of carbon monoxide for every one mole of carbon monoxide so what this is what we need for a reaction to occur this is what the balanced equation tells us it tells us that we need two moles of hydrogen for every mole of carbon dioxide based on what they've given us we just figured out that we have 2.56 moles of hydrogen for every mole of carbon dioxide so we have more than enough hydrogen right we only need two for every mole of carbon dioxide we have 2.56 so we have an excessive amount of hydrogen so the excess reactant is the hydrogen hydrogen is excess excess reactant and the other one is going to be the limiting reactant we don't we don't have enough carbon monoxide to react all of the hydrogen right we only have one for every 2.56 we would actually for this says you need one point you know two five or whatever for every uh one or one point two eight for every two point five six it should be a one to two ratio so we don't have enough carbon monoxide to react all of the hydrogen so carbon monoxide carbon monoxide is limiting limiting reactant now given that this is the excess reactant let's see how this the we can use the stoichiometric ratios to figure out how much methanol is going to produce it's all going to be limited by our by our carbon monoxide did i just say this is this is hydrogen is not the limiting reactant carbon monoxide is a limiting reaction we have more than enough hydrogen so how much carbon monoxide do we have we already figured it out we have 12.7 moles of carbon monoxide and we look at our we look at our let me write this over here so we have 12.7 moles of carbon monoxide and looking at our original equation we see for every mole of carbon monoxide we produce one mole of methanol so let's write that down times we want the the carbon monoxide in the denominator so for every one mole of carbon monoxide used we have one mole of methanol which is what c h three o h did i get that right yep produced we get that straight from the balanced equation and this is this math is pretty easy but it gives us the right units if we're going to be remember we're using the carbon monoxide not the hydrogen because the carbon monoxide is the limiting reagent that's what's telling us what's going you know if we can't if we use hydrogen as the limiting reactant then we wouldn't have enough carbon monoxide for the reaction to occur so this is what's kind of capping off on how much this reaction can move forward but the whole point of this was to cancel that and that so obviously for if we if we're using 12.7 moles of carbon monoxide we're going to produce 12.7 moles 12.7 moles of methanol will be produced will be produced and now we just have to figure out how what is the mass of 12.7 moles of methanol and we just just think about what the atomic weight so if you look at methanol ch3 let me put the h3 there h3oh its atomic weight is 12 plus 3 times 1 plus 3 plus 16 plus 1. so what is this this is 20 plus 2 is equal to 32 or we could say that if we think in molar terms or not in molar terms i should just say 32 moles this is its atomic this is its atomic weight sorry atomic weight so that tells us that if we have a mole of it we're going to have 32 grams of methanol per one mole for one mole of methanol and once again we got that by figuring out its atomic weight now to convert the number of moles of we have of methanol to the number of grams we just multiply that times that the units work out right this is the numerator this is in the denominator let me just copy and paste it so we have that copy and paste and then we can multiply it times that let me copy and paste it you get that right there and i'll pick a different color maybe a blue and just like that and then let's multiply these two we have moles of ca of methanol moles of methanol moles of methanol and we're left with 12.7 times 32 is equal to 400 actually i should just stay with significant so we'll say 406 grams i'm rounding down 406 grams of methanol let me write it with the units so grams i could grams of methanol went off the screen of methanol produced i could write the produced down here since we're all done so i think that was the first part of our question how many grams of methanol what mass of methanol that's 400 and i have a horrible memory 406 406 grams and then they say what mass of excess reactant remains after the limiting reactant has been consumed now there's a couple of ways you can do this the easiest way the easiest way to think about it is that the mass has to be conserved so we started off with we started off over here with 65 grams of let me be careful we started off here with 356 grams of carbon monoxide and 65 grams of hydrogen and we were able to produce 406 grams of methanol 406 grams and we figured out that carbon monoxide was the limiting reactant so all of this all of this gets consumed and only some of this gets consumed so if you do the math what gets consumed what gets consumed has to be equal to 406 grams because that's what gets produced so let's think about it a little bit what the left-hand side what how many total grams do we have so if we have 356 plus 65 we're starting off with 421 grams of reactant so we're starting with 421 grams of reactant 421 grams of reactant and then we end up with 406 grams 406 grams of product of our methanol so that means that 21 421 minus 406 grams of reactant was not used so that means that 421 minus 406 which is equal to what 21 minus 6 is 15 grams of reactant not not used now the reactant that's not going to be used is the one that you have an excess of and we figured it out that we have an excess of hydrogen we have an excess of hydrogen so all of that all of that 15 grams must have been 15 grams of hydrogen that was not used so 15 grams of hydrogen left over now the other way you could have done that the other way you could have done this exact same problem is you could have said look we we're starting with 12.7 moles of carbon monoxide that's the limiting reactant it's a two to one ratio you need two moles of hydrogen for every mole of carbon monoxide so you said okay if we have 12.7 moles of this i need twice that many moles of hydrogen so you would say well what's twice that that's 20 that's 25.4 you need 25.4 moles you have 32.5 so you'd subtract the difference you subtract 32.5 minus 25.4 and that number of moles of hydrogen is left over you'd multiply it times the the the grams per mole which is two and then you once again would get the same thing you would get 15 grams let's do that i just want to make sure you understand so we're starting off with we are starting off with 12.7 12.7 moles of carbon monoxide and we know and we know that we are required that two moles of hydrogen are required for every one mole of co of carbon monoxide that's required that is required so you know that this cancels with that you know 12.7 times 2 is 25.4 moles of h2 required and what is the mass of 25.4 moles of h2 so let's write that 25.4 moles of hydrogen required times well we know the molecular weight of mole of h2 is two so one mole we're going to have two grams of h2 per 1 mole of h2 this cancels out mole of h2 mole of h2 so we're going to have 25.4 times 2 is what that is 50.8 that is going to be 50.8 grams of h2 of h2 required required that's how much we're going to consume in the reaction now they wanted to know how much is left over so we're going to consume 50.8 we started with 65 so if you subtract 50.8 from 65 65 minus 50.8 you're going to get what you're going to get 14 point you're going to get 14.2 grams left over and that compares that compares to the 16 we got left over when we just took the 421 minus the 406 and the difference between the 14.2 and the 15 is really just a little rounding here and there with our digits anyway hopefully you found that helpful