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

• 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}