Question

If a solution containing 51.429 g of mercury(ii) perchlorate is allowed to react completely with a solution containing 16.642 g of sodium sulfate, how many grams of solid precipitate will be formed?

Answer 1

Hg(No3)2 +NaSO4 --->2NaNO3 + HgSO4(s)
calculate the moles of each reactant
moles=mass/molar mass

moles of Hg(NO3)2= 51.429g/ 324.6 g/mol(molar mass of Hg(NO3)2)=0.158 moles

moles Na2SO4 16.642g/142g/mol= 0.117 moles of Na2SO4

Na2SO4 is the limiting reagent in the equation and by use mole ratio Na2So4 to HgSO4 is 1:1 therefore the moles of HgSO4 =0.117 moles

mass of HgSO4=moles x molar mass of HgSo4= 0.117 g x 303.6g/mol= 35.5212 grams