Consider the balanced chemical equation: A2 + 2B2 → 2AB2. A react...

Question

Consider the balanced chemical equation: A2 + 2B2 → 2AB2. A reaction is performed with the initial amounts of A2 and B2 shown in part (a). The amount of product obtained is shown in part (b). Calculate the percent yield.a.

Accepted Answer

Welcome back everyone. The reactant below combined. According to the following equation where M two reacts with three moles of X two to form two moles of MX three, the initial amounts of M two and X two are shown in figure A where are gray molecules are our and molecules and our orange molecules are our X molecules. While the actual yield is shown in figure B, what is the percent yield of the reaction? So counting our molds of our reactant, we can count seven moles of our M two di atomic molecule and 15 moles of our ex to die atomic molecule. For figure B, we can count a total of seven moles of our M X three product as our actual yield. We're going to begin by calculating the theoretical yield of MX three from each reactant. So beginning with our MX three from our first reactant em too, we're going to begin with our seven molecules of M two multiplied by the ratio between molecules of M two and our denominator. And I'll use shorthand for that word, two molecules of MX three. In our numerator, we are going to refer to our given equation in the prompt where we see that we have for one molecule of M two reacted to molecules of MX three produced. So canceling out molecules of em to our reactant. We're left with molecules of our product MX three. And this is going to result in a value of molecules Of MX three that is formed from this first reactant. Now, we need to compare our theoretical yields of MX three from our second reactant X two. So we have 15 molecules of X two in which we want to multiply by the conversion factor to go from molecules of X two in the denominator two molecules of our product MX three in the numerator. Again, referring to our bounced equation given in the prompt, we have three molecules of X two reacted for every two molecules of MX three produced. So three molecules of X two and the dominator 22 molecules of MX three in the numerator canceling out molecules of X two were left with molecules of MX three. And this is going to result in three or sorry, rather 10 molecules of our product MX three that is formed from our second reactant. So we need to find the percent yield for our final answer. Recall that we can calculate percent yield By taking our actual yield divided by our theoretical yield times 100%. And so what we'll find is that our actual yield given from figure being our prompt, we determined was seven molecules of our product. So seven molecules of MX three in our numerator for the actual yield. And in our denominator are theoretical yield is going to be the reactant that produced the lower amount of our product, which is our limiting reactant. And because we have only 10 molecules of MX three formed from X two, that means that X two is our limiting reactant and our molecules of MX three formed is our theoretical yield. So we would plug that in our denominator 10 molecules of MX three as our theoretical yield. Now multiplying this quotient by 100%, we're going to come up with our percent yield. which in the next line is equal to a value of 70%. And this would be our final answer as our percent yield for the reaction producing MX three, the product which corresponds to choice, seeing the multiple choice as the correct answer. So I hope everything I went through is clear and let us know if you have any questions.

Struggling with GOB Chemistry?

Consider the balanced chemical equation: A2 + 2B2 → 2AB2. A reaction is performed with the initial amounts of A2 and B2 shown in part (a). The amount of product obtained is shown in part (b). Calculate the percent yield.a.

Watch next

Percent Yield practice set