Based on experiment 1:
Mass of Hg = 1.00 g
Mass of sulfide = 1.16 g
Mass of sulfur = 1.16 - 1.00 = 0.16 g
# moles of Hg = 1 g/200 gmol-1 = 0.005 moles
# moles of S = 0.16/32 gmol-1 = 0.005 moles
The Hg:S ratio is 1:1, hence the sulfide is HgS
Based on experiment 2:
Mass of Hg taken = 1.56 g
# moles of Hg = 1.56/200 = 0.0078
Mass of S taken = 1.02 g
# moles of S = 1.02/32 = 0.0319
Hence the limiting reagent is Hg
# moles of Hg reacted = # moles of HgS formed = 0.0078 moles
Molar mass of HgS = 232 g/mol
Therefore, mass of HgS formed = 0.0078 * 232 = 1.809 g = 1.81 g