) What is the volume of oxygen gas at STP from the decomposition of 10.8 g of mercuric oxide

(216.59 g/mol)?
__HgO(s) → __Hg(l) + __O2(g

*PLEASE SHOW STEPS

Not too bad. Take it slowly.

Balance the equation first:

__HgO(s) → __Hg(l) + __O2(g)
2HgO(s) → 2Hg(l) + O2(g)

Now, turn the 10.8 g of mercuric oxide into moles.

10.8 g HgO (1 mol HgO / 216.59 g HgO) = 0.0499 mol HgO

After that, relate the moles to the reaction. For every two moles of mercuric oxide decomposed, one mole of oxygen gas is produced.

0.0499 mol HgO (1 mol O2 / 2 mol HgO) = 0.02495 mol O2

Now, assuming STP (1 atm, 273K), find V in this equation (keeping in mind that n is equal to the number of moles of oxygen):

PV = nRT
(1 atm)V = (0.02495 mol O2)(0.082056 L*atm/mol*K)(273 K)
V = 0.559 L O2 (rounded to three significant figures)

Cheers!