How many grams of \textrm{Ag}_2 \textrm{O} are required to produce [var:O2] L oxygen gas via decomposition at 102.3 kPa and 175°C? 2

Question

2Ag2O=4Ag+O2

Answer 1

As pointed out by @Gabrielle M, we need to know the volume of O2 desired. For the sake of demonstrating how to approach this problem, I'll assume a volume of O2 of 1 liter of O2. Then we proceed as follows:

2Ag₂O ==> 4Ag + O₂ .. balanced equation

Use ideal gas law to find moles O2 and then use stoichiometry to find moles of Ag₂O. Finally convert moles Ag₂O to grams.

PV = nRT

P = pressure = 102.3 kPa

V = volume = 1 L

n = moles = ?

R = gas constant = 8.314 L-kPa/Kmol

T = temperature in Kelvin = 175C + 273 = 448K

Solve for n (moles of O2)

n = PV/RT = (102.3)(1) / (8.314)(448)

n = 0.0275 mols O₂

Use stoichiometry to find moles of Ag₂O needed:

0.0275 mols O₂ x 2 mol Ag₂O / mol O₂ = 0.055 mols Ag₂O needed

Use molar mass of Ag₂O (232 g/mol) to convert to grams:

0.055 mols Ag₂O x 232 g / mol = 12.8 g Ag₂O needed to produce 1 liter of O₂