In one experiment, the reaction of 1.00 g mercury and an excess of sulfur yielded 1.16 g of a sulfide of mercury as the sole product. In a second experiment, the same sulfide was produced in the reaction of 1.56 g mercury and 1.02 g sulfur. What mass of the sulfide of mercury was produced in the second experiment?

Question

Keesha

Chemistry

3 March, 04:44

In one experiment, the reaction of 1.00 g mercury and an excess of sulfur yielded 1.16 g of a sulfide of mercury as the sole product. In a second experiment, the same sulfide was produced in the reaction of 1.56 g mercury and 1.02 g sulfur. What mass of the sulfide of mercury was produced in the second experiment?

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Conway · Answer 1

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