Question

In the reaction, 2HgO(s)⟶2Hg(s)+O2(g) what mass of mercury(II) oxide, HgO, would be required to produce 919 L of oxygen, O2, measured at STP?

Answer 1

Molar volume at STP is 22.4 L/mol

Now use:
Mol of O2 formed = volume of O2 / molar volume
= 919 L / 22.4 L/mol
= 41.03 mol

From reaction,
Mol of HgO reacted = 2*mol of O2 formed
= 2*41.03 mol
= 82.06 mol

Molar mass of HgO,
MM = 1*MM(Hg) + 1*MM(O)
= 1*200.6 + 1*16.0
= 216.6 g/mol

use:
mass of HgO,
m = number of mol * molar mass
= 82.06 mol * 2.166*10^2 g/mol
= 1.777*10^4 g
Answer: 1.78*10^4 g