Question

If a solution containing 84.363g of mercury(II) Chlorate is allowed to react completely with a solution containing 14.334g of sodium sulfide, how many grams of solid precipitate will be formed? How many grams of the reactant in excess will remain after the reaction?

Answer 1

First the balanced chemical equation

Hg(ClO₃)₂ + Na₂SO₄ ==> HgSO₄ + 2NaClO₃

We see that the mercury chlorate and sodium sulfate react in a 1:1 mole ratio as the number in front of both formulae is 1.

We find the moles of mercury chlorate: its molar mass is 367.49 g/mol and we have 84.363 grams of it so the moles are mass/molar mass = 84.363 /367.49 = 0.23 mol.

For sodium sulfate the molar mass is 142.04 g/mol and we have 14.334 g of it so the moles of it are 14.334 g/142.04 g/mol = 0.101 mol.

Because the two react in a 1:1 ratio and there are more moles of Hg(ClO₃)₂ than Na₂SO₄, the chlorate is in excess and the sulfate is the limiting reagent, so all of it reacts, and only 0.101 mol of the 0.23 mol of chlorate will react with it to form 0.109mol of solid mercury(II) sulfate (as it is also in a 1:1 mole ratio with mercury chlorate and mercury sulfate).

So, mass of HgSO₄ formed = moles x molar mass ( which for mercury sulfate is 296.653 g/mol) so 0.101mol x 296.653 g/mol = 29.962 g.

We said the mercury chlorate will be in excess, by (0.23 - 0.101)moles = 0.129 moles.

Which translates to a mass of 0.129 mol x 367.49 g/mol = 47.41 g