How many molecules of H2S are required to form 79.0 g of sulfur according to the following resaction? Assume excess SO2.

Question

2 H2S(g) + SO2(g) -----> 3 S(s) + 2 H2O(l)

Answer 1

The stoichiometry is: (79.0 g S) * (1 mole S /32.06 g S) * (2 mole H₂S /3 mole S) * (6.022 x 10²³ molecules /1 mole H₂S) = 9.89 x 10²³ molecules of H₂S.