For the reaction shown, calculate the theoretical yield of the product (in grams) for each initial amount of reactants. Ti(s) + 2 F₂( g) → TiF₄(s) c. 0.233 g Ti, 0.288 g F₂

Video transcript

All right. Hi, everyone. So for this question, let's go ahead and calculate the theoretical yield of the product in grams. For each initial amount of reactants, we have 2.44 g of titanium and 8.82 g of co two or the reaction shut here, one mole of solid titanium combines with two moles of chlorine gas to produce one mole of T I cl four as a solid. Now, in this case, we're being told to convert each initial amount of reactants into the theoretical yield of the product. However, the true theoretical yield is based on which of the two is the limiting reagent because recall that the limiting reagent is going to end the reaction prematurely because it is the reagent that it gets consumed first. Whereas the excess reagent is always going to have some left over. Now, for this question, using dimensional analysis, we're going to use the mole ratios in the balanced equation provided and the molar masses of all compounds involved to relate the mass of each reactant to the theoretical mass of the product. But we're going to have to compare both values that we get because the theoretical yield depends on which of the two reagents is limiting. Meaning we're going to have to compare which reagent produces less product. So first, a few things to recall here, the molar mass of elemental titanium is 47 point 867 grams. Ple the molar mass of CO two is equal to 70.906 g per more. And the molar mass of T IC four is equal to 189.679 g per mole. So first, let's use the mass of titanium to calculate its theoretical yield of TCL four. So first, we have 2.44 g of titanium. So first, let's use the molar mass to convert grams of titanium into moles of titanium. So here we're going to multiply the mass of titanium by one mole of titanium over 47.867 g. Now placing grams of titanium in the denominator of the conversion factor ensures that grams of titanium cancel out. So now we can use the mole ratio to relate the moles of titanium to the moles of the product. So one more of T four is produced for every one mole of titanium consumed. And here moles of titanium cancel out. So now let's use the molar mass of TCO four to find the mass of product expected to be produced from 2.44 g of titanium. So that's 189.679 g over one mo So now after evaluating this expression, multiplying all numerators and dividing all denominators, our answer is equal to 9.68 g of T CO two. But before we can decide that this is going to be our final answer, we have to compare this value to the mass of product produced from the other reagent. So using CL two, now we have 8.82 g of C two, starting off with the molar mass of sealed two, that's one more oc two, over 70.906 g. So now now that we have our answer in units of moles of seal too, we can use the balanced chemical reaction to relate the moles of chlorine to the moles of product. So that's one mole of T I four divided by two moles of CO two. And so now we can use the molar mass of TC four to find grams. So that's 189.679 g of T IC L four divided by one more of T I four. Once again, making sure that my original units cancel out. And so the second quantity based off of the mass of C two is equal to 11.8 g of TC two. This is after rounding to three significant figures in both cases. Now, in this case, we can see that titanium produced the smaller amount of product. This means that titanium is the limiting reagent while chlorine is in excess. And so the theoretical yield for this reaction is based off the mass of titanium used since titanium is expected to be consumed completely during the reaction. So our theoretical yield is 9.68 g. But I just realized that I wrote TCL two and not T I four. So let me fix that. There we go. But going back to the very top of the screen, our answer is equal to option B in the multiple choice. That's 9.68 g of T IC L four and there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.