If a solution containing 59.91 g of mercury(II) nitrate is allowed to react completely with a solution containing 16.642
Posted: Thu Jun 09, 2022 10:33 am
If a solution containing 59.91 g of mercury(II) nitrate is allowed to react completely with a solution containing 16.642 g of sodium sulfate according to the equation below. Hg(NO3)₂(aq) + Na₂SO4 (aq) → 2 NaNO3(aq) + HgSO4(s) How many grams of solid precipitate will be formed? mass: How many grams of the reactant in excess will remain after the reaction? mass: g