If a solution containing 22.90 g of mercury(II) nitrate is allowed to react completely with a solution containing 7.410 g of sodium sulfate,
how many grams of solid precipitate will be formed? How many grams of the reactant in excess will remain after the reaction?
Balanced equation of the process is
Hg(NO3)2 + Na2S -----------> 2 NaNO3 + HgS
moles Hg(NO3)2 = 22.90 g / 324.61 g/mol = 0.0705 moles
moles of Na2S = 7.410 g / 78.04 g/mol = 0.0949 moles
Usig the mole ratio between Hg(NO3)2 + Na2S to find the limiting
reagent. The ratio is 1 to 1.
We have 1.3 times of Na2S as for Hg(NO3)2.
Hg(NO3)2 is therefore the limiting reagent. It will get used up
first then the reaction stops.
The ratio of Hg(NO3)2 to HgS is also 1:1
therefore moles of HgS produced = 0.0705 moles
mass of HgS produced = 0.0705 moles * 232.66g/mol
= 16.4025 g
the moles of Na2S remaining = 0.0244 moles
mass of Na2S remaining = 0.0244 moles * 78.04 g/mol
= 1.9041g
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