In this reaction, 5.02 L of oxygen gas were formed at 752 mm Hg and 27.3°C. How many grams of Ag2O decomposed? Ag2O(s) ==> Ag(s) + O2(g) (put answer in sig fig)

Answer 1

Convert mm Hg to atm: 752 mm Hg x 1 atm / 760 mm Hg = 0.989 atm

Convert 27.3ºC to Kelvin: 27.3 + 273.15 = 300.5K

Use the ideal gas law to find the number of moles (n) of O₂ present under these conditions:

PV = nRT

n = PV/RT = (0.989 atm)(5.02 L) / (0.0821 Latm/Kmol)(300.5)

n = 0.2012 mols O₂ present

Looking at the balanced equation for the decomposition of Ag₂O, we have..

2Ag₂O ==> 4Ag + O₂

So, 1 mol O₂ is equal to 2 mols Ag₂O

mols Ag₂O present = 0.2012 mols O₂ x 2 mols Ag₂O / mol O₂ = 0.4025 mols Ag₂O

mass Ag₂O decomposed = 0.4025 mols x 231.7 g/mol = 93.25 g = 93.3 g Ag₂O (3 sig. figs.)