Question

If a solution containing 115.34 g of mercury(II) Perchchlorate is allowed to react completely witha solution containing 16.642 of sodium sulfide, how many grams of solid ppt will be formed? -49.56

How many grams of the reactant in excess will remain after this reaction?

Answer 1

The molar masses of mercury(II) Perchchlorate and sodium sulfide are 399.5 g/mol and 78 g/mol respectively.

115.34 g of mercury(II) Perchchlorate corresponds to
16.642 of sodium sulfide corresponds to
Thus, mercury(II) Perchchlorate is excess reagent and sodium sulfide is the limiting reagent.
The moles of mercury(II) Perchchlorate that will remain in excess
Mass of mercury(II) Perchchlorate that will remain in excess

The number of moles of HgS formed = 0.2133
The molar mass of HgS is
Mass of HgS formed