You were going great.
n = 0.0154 mol Ag2O.
2Ag2O ==> 4Ag + O2
n O2 = 0.0154 mols.
Convert mols O2 to mols Ag2O.
2*0.0154 = 0.0308 mols Ag.
g Ag = mols x atomic mass.
%Ag = (g Ag/mass sample)*100 = ?
An 8.07g sample of impure Ag2O decomposes into solid silver and O2(g). If 395mL O2 is collected over water at 25C and 749.2mmHg barometric pressure, then what is the percent by mass of AgO2 in the sample? The vapor pressure of water at 25C is 23.8mmHg.
Could someone help me step by step with the question? So far this is what I have worked out:
Partial pressure of O2= .95atm
n of O2= .0154mol
I'm not sure where to go from here.
1 answer
1 answer