In the reaction, 2HgO(s)⟶2Hg(s)+O2(g) what mass of mercury(II) oxide, HgO, would be required to produce 905 L of oxygen, O2, measured at STP?

Answer 1

2HgO(s)==> 2Hg(s) + O₂(g) .. balanced equation

To find the mass of HgO, we will first find the moles of O₂ produced. Then using stoichiometry, we'll find moles of HgO needed and finally covert that to mass using the molar mass of HgO (216.6 g / mol)

At STP, one mole of any ideal gas = 22.4 L

905 L O₂ x 1 mol O₂ / 22. 4 L = 40.40 moles O₂ are present

Using stoichiometry and dimensional analysis, we find moles HgO:

40.40 mol O₂ x 2 mol HgO / 1 mol O₂ = 80.80 mol HgO

Converting mol HgO to g HgO:

80.80 mol Hg x 216.6 g / mol = 17502 g = 17,500 g (3 sig. figs.)