# If a solution containing 45 g of mercury(II) nitrate is allowed to react completely with a...

## Question:

If a solution containing 45.000 g of mercury(II) nitrate is allowed to react completely with a solution containing 14.334 g of sodium sulfate.

a) How many grams of solid precipitate will be formed?

b) How many grams of the reactant in excess will remain after the reaction?

## Precipitation Reactions:

Precipitation reactions are a type of double-displacement reaction (metathesis) where the result is a solid precipitate forming from the mixing of two aqueous solutions. The reason the precipitate forms is the resulting product compound has limited or no solubility in the given solvent and, therefore, forms a solid. Understanding the ionic compound solubility rules/guidelines is one way of judging whether a precipitate will form and what the possible precipitate might be.

• a) There will be 30.0 grams of mercury(II) sulfate precipitate formed from this reaction.
• b) There will be 12.3 grams of unreacted mercury(II) nitrate left in solution.

> a) The best way to start is to write a balanced reaction equation for this reaction:

{eq}\rm{Hg(NO_3)_2(aq)\:+\:Na_2SO_4(aq)\:\rightleftharpoons \:2NaNO_3(aq)\:+\:HgSO_4(s)} {/eq}

Next we need to know which compound is the limiting reagent which will determine the amount of product formed. We will convert the masses to moles using their respective molar masses and then see which one will produce less moles of product:

{eq}\rm{(45.000\:g\:Hg(NO_3)_2)\left(\dfrac{1\:mole\:Hg(NO_3)_2}{324.62\:g\:Hg(NO_3)_2}\right)\left(\dfrac{1\:mole\:HgSO_4}{1\:mole\:Hg(NO_3)_2}\right)\:=\:0.139\:moles\:HgSO_4}\\ \rm{(14.334\:g\:Na_2SO_4)\left(\dfrac{1\:mole\:Na_2SO_4}{142.04\:g\:Na_2SO_4}\right)\left(\dfrac{1\:mole\:HgSO_4}{1\:mole\:Na_2SO_4}\right)\:=\:0.101\:moles\:HgSO_4} {/eq}

This tells us that the sodium sulfate is the limiting reagent as we can only make 0.101 moles of mercury(II) sulfate as opposed to 0.139 moles from mercury(II) nitrate. We now convert the moles of mercury(II) sulfate we can make to grams using its molar mass:

{eq}\rm{(0.101\:moles\:HgSO_4)\left(\dfrac{296.65\:g\:HgSO_4}{mole\:HgSO_4}\right)\:=\:30.0\:g\:HgSO_4} {/eq}

> b) From here we need to convert the difference of product we can theoretically make from the sodium sulfate and what we can theoretically make from mercury(II) nitrate to mass of mercury(II) nitrate:

{eq}\rm{0.139\:moles\:HgSO_4\:-\:0.101\:moles\:HgSO_4\:=\:0.038\:moles\:HgSO_4}\\ \rm{(0.038\:moles\:HgSO_4)\left(\dfrac{1\:mole\:Hg(NO_3)_2}{1\:mole\:HgSO_4}\right)\left(\dfrac{324.62\:g\:Hg(NO_3)_2}{1\:mole\:Hg(NO_3)_2}\right)\:=\:12.3\:g\:Hg(NO_3)_2} {/eq}