Hydrogen sulfide is an impurity in natural gas that must be removed. One common removal method is called the Claus process, which relies on the reaction: 8 H2S1g2 + 4 O21g2¡S81l2 + 8 H2O1g2 Under optimal conditions the Claus process gives 98% yield of S8 from H2S. If you started with 30.0 g of H2S and 50.0 g of O2, how many grams of S8 would be produced, assuming 98% yield?

Video transcript

Hello everyone today, we are being given the following question and asked to solve for the mass of selenium oxide. So we have the following reaction that has a yield of 96%. We have a balanced chemical equation here We are being asked to determine the mass of selenium oxide produced if 20 g of hydrogen selenium And 40 g of oxygen gas were reacted together. So the first thing I want to do is find the limiting reactant. And to do that we're gonna take each Amount of reaction that we have and convert it into most of a product. And in this case we're going to the product of selenium oxide. So we're gonna take our 20 g of H two s. c. And to convert it to moles of our product. We first want to do the molar mass. So we're gonna say one mole of H two S. E. Is equal to its molar mass, which is 80.98 grounds H two s. E. And then since we want to get moles of our products, lithium oxide, we have to do the multiple ratio and we do that by using the coefficients that are in front of each reactant or product. So we have two moles of H two S. E. And we put that in the denominator so that our units can cancel later on. And for every two moles of H two S. C. We produce two moles of S. E. 02 or selenium oxide. Our units cancel as such. And we're left with a value of 0. malls of selenium oxide. We're gonna do the same process with the oxygen gas. We're gonna take our 40 g of 02. We're gonna do our mole mass. We're going to say one mole of 02 Is 32 g. And then we're gonna do the multiple ratio once again. But instead we're going to say we have three moles of 02. Right? Using that coefficient and that's going to produce two moles of psyllium oxide. Our units cancel once again and we are left with 0.833 moles of sodium oxide. Now limiting reaction is the reaction that produces the least amount of product. And in that case that is going to be our H two S. E. So we're gonna make a note here, H two SC. Is our limiting reactant. Our 2nd step is using that Mole using the moles of our limited reactant and converting that into grams of our product. So we're going to say 0.2470 moles. We're going to use the molar mass for this, we're going to say one mole of S. C. 02 Is equal to the Molar mass which is 110.96 g. Our units cancel. And we're left with a value of 27.40 g of selenium oxide. Our third and final step is going to be solving for the actual percent yield that is going to be the actual yield Over times the theoretical yield, I was gonna say theoretical, That gives us our actual which is 96%. So they're going to say 96 over 100 times our theoretical, which is 27.40 grams of selenium oxide, Which when we do the math yields us 26. grams of selenium oxide. As our final answer, I hope this helped, and until next time.