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