posted by Afitz on .
What is the maximum mass of S8 that can be produced by combining 80.0 grams of each reactant?
8SO2 + 16H2S -> 3S2 + 16H2O
you have to figure how many moles you have of SO2:80/64
of H2S: 80/34
You need twice as much of H2S as SO2.
If you have more than twices as much, then the limiting reactant is SO2. for S8 (you have S2), it will be then 3/8 of the moles you have of SO2.
Now if you have less than twice as much, then the limiting reactant is H2S, and the moles of S8 will be 3/16 of the amount of H2S you had.
Someone answer the damn question already!
144 g s8
ha ha ha ha ha
It's actually 113 according to my book.