For this reaction we have molecular hydrogen which is different from elemental hydrogen, and we have carbon monoxide which is a compound formed from two elements carbon and oxygen.īut I get what you're asking, what if only know the mass of a single reactant and not both? Which one would be the limiting reactant then if we don't know how much of one reactant we have? Usually if only a single reactant's mass is given it is because they specify that that other reactant is added in excess or we have an unlimited supply of it. I believe you mean chemical species instead of elements because we can have a compound with multiple elements in them which we are given mass information about. Specifically here we just have to add that ratio of products to reactants step in the middle of the conversions if the ratio wasn't 1:1. This works for any situation where you have to convert multiple times. Finally we multiply the moles of the product by the molar mass to get the grams of our product. Then multiply those grams by 2:4 which is the ratio of products to reactants to get moles of product. So we see that if we divide our original grams of reactant by the molar mass, we get moles of our reactant. It also gets rid of any doubt whether we have to divide or multiply in a step. So if you follow it you can see each unit diagonal to the same of its kind is cross canceled each time we multiply or divide to get a new desired unit. The math would look as follows: (reactant grams/1) x (1 mol reactant/ reactant grams) x (2 mol product/4 mol reactant) x (product grams/1 mol product). And then we could also find the grams of the product if they asked us to by using the molar mass of the product. The idea would be to turn grams of the reactant into moles of the reactant using the molar mass of the reactant, then multiply by the mole-to-mole ratio in the balanced chemical equation to get moles of your product. Where the reactant has a coefficient of 4 and the product has one of 2. Using your example, let's say they gave us grams of a reactant (and we assume we already know it's the limiting reactant) and want us to find the theoretical yield of a product. You cross cancel units like you would in fractions to get the units you desire. It's part of dimensional analysis which lets you do successive conversions like this by either multiplying or dividing. Hope that helps.Īfter you've turned the grams of the reactants into moles of reactants and have found the limiting reactant, you would multiply by the mole-to-mole ratio. If we had done it the other way and multiplied the 32.2 mols of H2 by 2 instead, that would tell us we need 64.4 mols of CO to react which doesn't make sense since we know we need more moles of H2 than mols of CO for the reaction to occur. So we multiply the CO mols by two to know how many moles of H2 gas would be required to react with the entirety of the CO supply because of that 1:2 ratio. But since we don't have 16.1 mols of CO, we have less than that, the CO is therefore the limiting reactant. If you wanted to use up the entire 32.2 mol supply of H2, you would need 1/2 of the 32.2 mols for the required mols of CO, or 16.1 mols of CO. So 12.7 mols of CO would require twice the amount of mols of H2, or 25.4 mols of H2. You need twice as much H2 as CO since their stoichiometric ratio is 1:2. For the CO if you were to use it up completely you would use up 12.7 mols of CO. Of the two reactants, the limiting reactant is going to be the reactant that will be used up entirely with none leftover. The previous replier is correct, just want to extra some extra bits.
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