Use PV = nRT and solve for n = moles oxygen.
Use stoichiometry to convert moles O2 to moles Ag2O, convert to grams Ag2O
In this reaction, 4.58 L of O2 were formed at 745mm Hg and 308 K. How many grams of Ag2O decomposed.
2 Ag2O----> 4 Ag + O2
Please help I'm very confused
2 answers
As Dr. Bob said:
Use PV=nRT to solve for n = moles of oxygen
Convert 745mmHg to 1 atm, thats 745mmHg x 1 atm/760mmHg = 0.98atm
(0.98)(4.58L)/(0.0821)(308K) = 0.18mol O2
Use Stoichiometry to convert O2 moles to moles Ag20, then convert that to grams Ag2O
0.18molO2 x 2molAg2O/1molO2 x 231.735gAg2O/1molAg2O = 83.4gAg2O
*the exact answer is 82.3g bc I rounded different then the book, but this is how you get to that number
Use PV=nRT to solve for n = moles of oxygen
Convert 745mmHg to 1 atm, thats 745mmHg x 1 atm/760mmHg = 0.98atm
(0.98)(4.58L)/(0.0821)(308K) = 0.18mol O2
Use Stoichiometry to convert O2 moles to moles Ag20, then convert that to grams Ag2O
0.18molO2 x 2molAg2O/1molO2 x 231.735gAg2O/1molAg2O = 83.4gAg2O
*the exact answer is 82.3g bc I rounded different then the book, but this is how you get to that number
1 answer