Hg2O(s) -> Hg(l) + O2 (g) (unbalanced)

Question

Hg2O(s) -> Hg(l) + O2 (g) (unbalanced) Calculate the volume of oxygen gas, measured at 30.0C and 725 torr that can be produced from complete decomposition of 1.92 grams of mercury(II) oxide. Answer has to be in Liters (L)

Answer 1

We must always balance the chemical equation first:

2HgO → 2Hg + O₂

Now, we simply apply stoichiometry and Ideal Gas Law:

1.92 g HgO x (1 mol HgO/216.59 g HgO) x (1 mol O₂/2 mol HgO) = number of moles, n

Ideal Gas Law:

PV = nRT

V = nRT/P

Now, let's plug in our stoichiometry above into the Ideal Gas Law, as well as all of our other parameters, and perform the proper dimensional analysis:

V = ((1.92 g HgO x (1 mol HgO/216.59 g HgO) x (1 mol O₂/2 mol HgO)) x (0.0821 L*atm/mol*K) x (30.0 + 273.15)K)/(725 torr x (1 atm/760 torr))

V = 0.116 L O₂