Question

Please explain questions are on finals and really need to understand the process.

If a solution containing 45.156 g of mercury(II) chlorate is allowed to react completely with a solution containing 14.334 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

The raection between sodium sulfate and mercury(II) chlorate is as follows

Hg(ClO3)2 + Na2SO4 ---> HgSO4 (s) + 2NaClO3

We see from the stoichiometry of the reaction that 1 mole of Hg(ClO3)2 reacts with 1 mole of Na2SO4 to produce 1 mole of HgSO4 (solid precipitate)

Given,

Mass of mercury(II) chlorate = 45.156 g

Mass of sodium sulfate = 14.334 g

We know that,

Molar Mass of mercury(II) chlorate = 367.49 g / mol

Molar mass of sodium sulfate = 142.04 g / mol

=> Moles of mercury(II) chlorate = 45.156 / 367.49 = 0.1229 moles

Moles of sodium sulfate = 14.334 / 142.04 = 0.1009 moles

The molar ration required according to the stoichiometry is 1:1 but we have less sodium sulfate as compared to mercury(II) chlorate

Therefore mercury(II) chlorate is in excess and remins

Moles of HgSO4 (solid precipitate) fromed = 0.1009 moles

Molar mass of HgSO4 = 296.653 g / mol

=> Mass of Solid precipitate formed = 0.1009 x 296.653 = 29.94 g

Moles of mercury(II) chlorate left = 0.1229 - 0.1009 = 0.022 moles

=> Mass of mercury(II) chlorate left = 0.022 x 367.49 = 8.08 g