Hg(NO3)2 + Na2SO4Â Â --------------------> HgSO4 + 2
NaNO3
324.6 g        142
g                                  Â
296.6Â Â Â Â Â Â Â 170 g
22.90Â Â Â Â Â Â Â Â Â Â Â Â
7.410Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
??
here limiting reagent is Na2SO4. so product formed according to
that.
142 g of Na2SO4 ----------------- 296.6 HgSO4
7.410 g of Na2SO4 ----------------- ??
mass of HgSO4 = 296.6 x 7.410 / 142
                        Â
= 15.48 g
mass of solid precipitate will be formed = 15.48
g
here Hg(NO3)2 is excess reagent. because
we need only 16.94 g of Hg(NO3)2. but we have 22.90 g so
the excess amount is = 22.90 - 16.94 = 5.96
g
