Iron(II) sulfide reacts with hydrochloric acid according to the reaction: FeS(s) + 2 HCl(aq) → FeCl₂(s) + H₂S(g) A reaction mixture initially contains 0.223 mol FeS and 0.652 mol HCl. Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant remains?

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Hi, everyone and welcome. Our next problem says iron two sulfide reacts with hydrochloric acid. According to the reaction, Fes solid plus two HCL, aqueous becomes fecl two solid plus H two S gas. A reaction mixture initially contains 0.223 moles of Fes and 0.652 moles of HCL. Once the reaction has occurred as completely as possible, what amount in moles of the extent excess reactant remains a 1.02 moles B 0.505 moles C 0.2206 moles or D 0.152 moles. So let's think through what's happening here and what we're being asked, we have a reaction and we're saying we're going to make it occur as completely as possible. So until one of our reactants is used up, so we need to identify which of these reactions is the limiting reactant. And then our question is asking us once we've used up that limiting reactant, how much is left over of the other one, the one we have in excess. Well, to do that, we need to know how much product will be produced from our limiting reactant. So then from this, from this, how much product is produced? And then finally, how much of our excess reagent did it take to react with that amount of limiting reactant? So how much of the excess reactant did this use up? So the first step before we can do any of this is do we have a balanced equation? Because we're going to need to look at this trom Mery here. So we have FES plus two HCL goes to FECL two plus H two S. So in terms of iron, we have one on the left, one on the right hydrogen, we have two on the left two on the right sulfur, we have one on the left, one on the right and chloride, we have two on the left two on the right. So yes, we have a balanced equation so we can use these coefficients. It's gonna erase the numbers I just wrote to convert between moles of different elements. So first, we have to figure out which reaction is limiting. So first, for Fes, we have 0.223 moles of Fes. And we want to say how much product will this gene will be produced? Well, we use our FECL two just because it's uh our solid product as opposed to our gas. And we know that we have one so multiplied by one mole of Fes. But wait a minute, we don't want to do that. We want to cancel out. So we need to e our conversion factor, our moles of FES needs to be on the bottom. So I'm drawing a line for my conversion factor that division sign and I'll put one mole fes on the bottom so that moles of FES will cancel out. And I know I generate one mole of F EC2. So that's the way to calculate how much product is produced. It's just multiplying by one. So 0.223 moles FC two, and then we need to do the same thing with our other reagent. So for HCL, we have 0.652 moles HCL multiplied by. Well, in our balance equation, we have two moles HCL. So we put that on the bottom of our conversion factor and that generates one mole of FECL two. So we'll put that on the top one mole FECL two. So we're essentially dividing our amount of HCL in half. And that result gives us 0.326 moles of CL two. So which is our limiting reactant, it will be the one that produces the least amount of product. And that is fes with its 0.223 moles. So this is our limiting reactant. So now we need to say once we've used up all of that, so we've produced 0.223 moles of F BC L two, how much HCL is left over? So I'm just going to scroll up a little bit to make more room for my calculations. Well, we'll see how many moles of hydrogen chloride does it take to produce that much product? So, we'll start with 0.223 moles of epic CL two. And again, set up our conversion factor for every one mole fecl two. We put that on the bottom to cancel out. We needed to have two moles of HCL. So essentially multiplying by two and that ends up giving us 0.446 moles of HCL needed. So when the fes is completely used up, we'll have consumed this much hydrochloric acid. So last of all, we just need to subtract the amount we have the amount we need from the amount we have. So we have 0.652 moles of HCL, subtract 0.446 moles of HCL need it. And the amount we have in excess is 0.206 moles of HCL left. So when we look at our answer choices, we see that choice C is 0.206 moles. So once again, to calculate, once the reaction has occurred as completely as possible, what amount in moles of the excess reactant remains? We calculated, we checked for a limiting reactant, calculated how much product will be generated and how much of our excess reactant was used up in that process. And then just subtracted that from how much we have getting our answer of choice. C 0.206 moles. See you in the next video.

Verified Solution

Key Concepts

Stoichiometry

Limiting Reactant

Excess Reactant

Iron(II) sulfide reacts with hydrochloric acid according to the reaction: FeS(s) + 2 HCl(aq) → FeCl2(s) + H2S(g) A reaction mixture initially contains 0.223 mol FeS and 0.652 mol HCl. Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant remains?

Verified Solution

Key Concepts

Stoichiometry

Limiting Reactant

Excess Reactant