What is the maximun mass of S8 that can be produced by combining 85.0 g of each reactant? 8SO2+16H2S yield to 3S8+16H2O

Question

What would the maximum mass be?

Answer 1

The limiting reagent of this reaction will be the reactant which provides the least amount of S8. Incidentally, this lower mass of S8 will the be the maximum amount you can synthesize based on the amounts given. I'll set up the stoichiometric conversion for one of them and I'll leave the rest for you to finish up:

 

(85.0 g SO2) x [(1 mole SO2)/(64.066 g SO2)] x [(3 moles S8)/(8 moles SO2)] x [(256.48 g S8)/(1 mole S8)] = 127.6 g S8.

 

Follow the same approach, starting from H2S, while making sure to use the right ratio of moles and molecular weight. Upon comparison with the answer above you'll know which is the limiting reagent and what the maximum mass of S8 can be.