Question

If you started with 20.0 grams of H2S and 50.0 grams of O2, how many grams of S8 would be produced, assuming 96% yield?

Answer 1

Can you come up with the balanced chemical equation?

Answer 2

Yes, one sec!

Answer 3

8H2S(g)+4O2(g)=S8(l)+8H2O(g)

Answer 4

you'd wanna find the limiting reactant now.

Answer 5

Under optimal conditions the Claus process gives 98% yield of S8 to H2S. Does that help at all?

Answer 6

you would find the limiting reactant by find the amount of moles of each reacting species and dividing it by it's stoichiometric coefficient (from the balanced equation).

Answer 7

by finding*

Answer 8

obviously, the one with less moles is the limiting factor.

Answer 9

I'm lost.

Answer 10

you would find the number of moles \(n=\dfrac{m}{M}\)

M= molar mass, m=mass (in grams), n= moles

then divide each by it's coefficient in the equation.

8H2S(g)+4O2(g)=S8(l)+8H2O(g)

e.g. \(\dfrac{n_{O_2}}{4}\)= number of "normalized" moles

i'm using the term "normalized" in a statistical sense.