Question

You have a badly tarnished silver teapot. The tarnish is a result of the silver metal reacting with the oxygen in the air according to

4Ag(s) + O2(g) = 2Ag2O(s),
Delta H = -62.2 kJ/mol


You carefully clean the teapot and then, out of curiosity, you weigh it. The teapot has a mass of 877 g. After some more measuring and calculating, you also determine that 6.70 % of the silver is on the surface and exposed to the air.

If the teapot were once again to become completely tarnished, how much mass (Delta m) is converted to energy?

E=mc^2, c= 3*10^8 m/s

Answer 1

I would multiply 877 x 0.067 = g Ag exposed.
Change to moles Ag, use the equation to convert to kJ, then use delta E = delta m*c^2 and calculate delta m. Post your work if you get stuck.

Answer 2

I got stuck.

Answer 3

877g * .0670 (the percent on surface) = 58.759 * 1molAg/107.9g = .5446 mol Ag

-62.2 kJ/mol /4 (from chem equation) =
15.55 kJ (can lose the negative sign)

-15.55kJ * .5446 mol Ag = -8.46853 kJ *1000J = 8468.53 J
8468.53J = m(3.0x10^8)^2 solve for m

answer will be in Kg so multiply by 1000 to get grams.