Question

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?

Answer 1

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