Question

10.4g of Mercury(2) oxide will produce what volume of oxygen in the reaction where mercury(1) oxide and oxygen produces mercury(2) oxide

Answer 1

Ans. 2 Hg₂O + O₂ -----------> 4 HgO

2 moles of mercury(I) oxide combines with one mole of O₂ to form 4 moles of mercury(II) oxide. This reaction consumes (but do not produce) oxygen.

Moles of HgO in 10.4 g sample = mass / molecular mass

                                                = 10.4 g / 216.59 g mol^-1

                                                = 0.048 moles

#1. Assume another reaction [wherein O₂ is produced as a byproduct]:

            Hg₂O + O₂ -----------> 2 HgO + (1/2) O₂

In above reaction, Hg₂O oxidation is carried out under controlled condition to get the stoichiometric reaction as mentioned.

Formation of 2 moles of HgO produces half moles of O₂.

So, moles of O₂ produced = ½ x moles of HgO produced

                                    = ½ x 0.048 = 0.024 moles

Volume of 1 mol gas at STP = 22.4 L mol^-1

Volume of 0.024 moles O₂ = 22.4 L mol^-1 x 0.024 moles = 0.5376 L

                                    = 537.6 mL

Thus, volume of O₂ produced = 537.6 mL = 0.5376 L