Posted by lee on .
If a solution containing 34.576 g of mercury(II) perchlorate is allowed to react completely with a solution containing 10.872 g of sodium sulfate, how many grams of solid precipitate will be formed?
Hg(ClO4)2 + Na2SO4 = HgSO4 + 2NaClO4
So, each mole of Hg(ClO4)2 requires one mole of Na2SO4
34.576g Hg(ClO4)2 = 34.576/399.49 = 0.0865 moles
10.872g Na2SO4 = 10.872/142.04 = 0.0765 moles
So, 0.0765 moles of HgSO4 and 0.1530 moles of NaClO4 are produced. Convert the moles to grams for the desired products for the answer.
If a solution containing 40.80 g of lead(II) nitrate is allowed to react completely with a solution containing 7.410 g of sodium sulfide, how many grams of solid precipitate will be formed?