Asked by Monique
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 Ag2O----> 4 Ag + O2
Please help I'm very confused
Answers
Answered by
DrBob222
Use PV = nRT and solve for n = moles oxygen.
Use stoichiometry to convert moles O2 to moles Ag2O, convert to grams Ag2O
Use stoichiometry to convert moles O2 to moles Ag2O, convert to grams Ag2O
Answered by
Migueal
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
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