A 0.550 g sample containing Ag2O and inert material is heated, causing the silver oxide to decompose according to the following equation: 2 Ag2O(s) → 4 Ag(s) + O2(g) If 13.8 mL of gas are collected over water at 27°C and 1.00 atm external pressure, what is the percentage of silver oxide in the sample? The partial pressure of water is 26.7 mm Hg at 27°C

This problem we will solve using the stoichiometry of the reaction since by using the ideal gas law we can calculate how many moles of O 2 were produced in the reaction. We then obtain the moles of Ag2O and convert it to grams and finally calculate the percentage of silver oxide in the sample.

PV = nRT ====== n = PV/RT

where P is the Pressure obtained by substracting the partial pressure of water from the 1 atm pressure

PH2O : 26.6 mmHg / 760 mmhg/atm : 0.034 atm

P: ( 1-0.034 ) atm : 0.966atm

T: (27+273) K : 300 K

R: 0.08205 LAtm/(molK)

V=: 13.8 mL/1000mL/L : 0.0138 L

n= 0.966 atm * 0.0138 L/(0.08205 LAtm/Kmol*300K) = 0.00054 mol

MW Ag2O = 231.6 g/mol

since 1 mol O2 is produced from 2 mol Ag2O per 0.00054 required 0.00108 mol Ag2O

converting to grams

0.00108 mol * 231.6 g/mol = 0.250 g of Ag2O present in the original sample.

0.250 g/0.550 g *100 = 45.5 %