Question

If a solution containing 24.02 g of mercury(II) acetate is allowed to react completely with a solution containing 6.256 g of sodium sulfate, how many grams of solid precipitate will be formed? How many grams of the reactant in excess will remain after the reaction?

Answer 1

Balanced chemical equation is;

Hg(OAc)2 + Na2SO4 --------------> HgSO4 (s) + 2 NaOAc

No. of moles of Hg(OAc)2 = 24.02 / 318.678 = 0.07537 moles

No. of moles of Na2SO4 = 6.256 / 142.035 = 0.044 moles

equal moles of Hg(OAc)2 and Na2SO4 will react to give equal moles of HgSO4 precipitate.

Here, limiting reagent is Na2SO4 (one which is present in limited quantity)

therefore, no. of moles of precipitate formed = 0.044 moles

'weight of HgSO4 precipitate formed = 0.044 x 296.65 = 13.066 grams of HgSO4 formed

excess Hg(OAc)2 remain is = 0.07537 - 0.044 = 0.03137 moles

Weight of Hg(OAc)2 remain = 0.03137 x 318.678 = 9.99 grams