Question

Calculate and state clearly the mass of Rh2S3 formed after 30.0 g rhodium with 30.0 g silver. State what the limiting reactant is and write an equation first.

Answer 1

    2Rh + 3S -----------> Rh2S3

no of moles of Rh   = W/G.A.Wt    = 30/103   = 0.3 moles

no of moles of S      = W/G.A.Wt    = 30/32      = 0.94 moles

3 moles of S react with 2 moles of Rh

0.94 moles of S react with = 2*0.94/3   = 0.63 moles of Rh is required

Rh is limitng reactant

2 moles of Rh react with S to gives 1 mole of Rh2S3

0.3 moles of Rh react with S to gives = 1*0.3/2   = 0.15 moles of Rh2S3

mass of Rh2S3    = no of moles * gram molar mass

                             = 0.15*302   = 45.3g of Rh2S3 is solid >>>>>answer

30g of silver is wrong 30g of sulfur