General Chemistry

Question

What is the maximum mass of S8 that can be produced by combining 77.0g of each reactant?

8SO2+16H2S-->3S8+16H2O

Answer: Number __ g S8

Answer 1

8SO2 + 16H2S ==> 3S8 + 16H2O ... balanced equation
Find the limiting reactant:
moles SO2 = 77.0 g x 1 mole/64 g = 1.20 moles
moles H2S = 77.0 g x 1 mole/34 g = 2.26 moles
Since it takes twice as much H2S as SO2 (16 v 8 in balanced eq.), H2S is the limiting reactant b/c there isn't twice as much.
Moles S8 formed = 2.26 moles H2S x 3 moles S8/16 moles H2S = 0.425 moles S8
mass of S8 formed = 0.425 moles x 256 g/mole = 109 grams SO2 formed (to 3 significant figures)