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

Question

Calculate the volume of oxygen gas, measured at 30.0C and 725 torr that can be produced from complete decomposition of 8.94 grams of mercury(II) oxide. Answer has to be in Liters (L)

Answer 1

2Hg₂O ==> 4Hg + O₂. ... balanced eq

moles Hg₂O used = 8.94 g x 1 mole/417 g = 0.0214 moles

Moles O₂ that can be produced = 0.0214 mol Hg₂O x 1 mol O₂/2 mol Hg₂O = 0.0107 mol O₂

Using the ideal gas law PV = nRT, calculate volume...

V = nRT/P = (0.0107 mol)(0.0821 Latm/Kmol)(303K)/0.954 atm

V = 0.279 L

NOTE: 725 torr was converted to atm using 1 atm = 760 torr. Also 30 deg C was converted to K by adding 273.