Question

If a solution containing 36.48 g of mercury(II) chlorate is allowed to react completely with a solution containing 6.256 g of sodium sulfide, how many grams of solid precipitate will be formed? And how many grams of the reactant in excess will remain after the reaction?

Answer 1

Balanced Chemical equation is:

HgCl2 + Na2S --------> HgS (precipitate) + 2 NaCl

from this 1 moles of each reactant will react to give 1 mole of precipitate HgS

no. of moles of HgCl2 = 36.48 / 271.49 = 0.13436 moles

no. of moles of Na2S = 6.256 / 78.04 = 0.080 moles

Here one which is present in limited quantity is the limiting reagent

so, limiting reagent is = Na2S

Na2S will react with excess HgCl2

the no. of moles of precipitate formed = 0.080 moles of HgS

Weight of HgS formed = 0.080 x 232.65 = 18.65 grams of HgS formed

excess reactant remain = 0.13436 - 0.080 = 0.05436 moles

weight of excess reagent = 0.05436 x 271.49 = 14.76 g