The diagram represents a mixture of AB2 and B2 before it reacts to form AB3. (Red spheres = A, blue spheres = B.) Which reactant is limiting, and how many AB3 molecules are formed? (LO 3.7)

(a) B2 is limiting, and 10 molecules of AB3 are formed. (b) B2 is limiting, and 4 molecules of AB3 are formed. (c) AB2 is limiting, and 6 molecules of AB3 are formed. (d) AB2 is limiting, and 4 molecules of AB3 are formed

Video transcript

Hello everyone. So in this video we're given this illustration right over here and basically it's just a mixture of a to B&B. two. So the question is asking us what is the limiting reagent and how much of the product is going to be formed? So first thing I'm gonna do is kind of translate this reaction over here into a chemical reaction. So of course we have our starting materials being A to B and B. Two And this forms a. b. three. And of course just like any chemical reaction, we have to balance this. So starting off, of course we just have materials of A. And B. Components as well as our product side. So on the starting material side right now we have three items of B. And then we have two atoms of A. On the product side we have one item of A. And three items of B. So the coefficient I'm imagining product side is by adding a coefficient of four. That gives us four days now and then four times three is 12. So we have 12 base. And to bounce these with this side. So the left side, the sort of material side, let's go ahead and actually add a coefficient of 22 R. A to B. This gives us now Wolf two times two is four. So we have four items of eight. And then we have two B's and two B's. So two plus two is four. And lastly of course we just need to balance out Arby's so well i coefficient of 52 R. B. Two, what does gets as well, five times two is 10. And then we have two of our bees. So 10 plus two is equal to 12. Now this chemical reaction is now properly balanced. Now, taking a look at this illustration here, let's circle our A to B. S in purple. So we have 12345 and six. So we have two moles of A to B in our actual solution. And then in red I'll go ahead and circle might be too. So which is just 12 and three. And then we have three moles of B2. Now, doing some dimension analysis, we can solve for our limiting reagent. Let's start off with our six moles. Our first study materials are A to B. So we can see from our chemical reaction over here. So for every two modes of A T. B, we get four moles of our product. So that's exactly what I'll use as my conversion factor. So two rules of a to B. And we get four moles of a B. Three. We see now that the moles of a TB will cancel Putting these numbers into my calculator. I get the value of 12 moles of a b. three. So 12 miles away. B three is going to be, how much product we get when we use the six moles of our first starting material. Let's do the exact same method of calculation for our second star material, that is three moles of a two or b to apologize. So three moles of B two. Again. Again. Look at my chemical reaction here, that's balanced for every five moles of B to get for most of our products. Again, for every five moles of our B two, we get four moles of a. B three. Again we see the moles of B two canceling. So once I put this into the calculator, I get the value of 12/5, which is just 12 or two 0.4. Who knows? So that's a decimal form of our a B three, our product. So we now compare the value of this and this here that this has less product and therefore making our B2 star material are limiting regions. So our answer is that our limited region, so L. R is equal to B two and we'll just round are 2.422. So two moles of a. B. Three. Our product worms. Alright, everybody. So this is going to be my final answer for this problem. Thank you all so much for watching