A digital watch battery draws 0.15 milliamperes of current, which is provided by

Question

A digital watch battery draws 0.15 milliamperes of current, which is provided by a mercury battery whose net reaction is:

HgO(s) + Zn(s) ? ZnO(s) + Hg(l)

If a partially used battery contains 1.80 g of each of these four substances, for how many hours will the watch continue to run?

(Determine the limiting reagent in terms of moles. Then multiply by the stoichiometric ratio to determine the number of moles of electrons produced. Then convert moles of electons into Coulombs (Amps*seconds). Then you just need to divide by amps used to find the total number of seconds, and convert to hours.)

Explanation / Answer

The molar masses of HgO and Zn are: 216.59 g/mol and 65.38 g/mol, respectively. Hence HgO is the limiting reactant, and it has:
(1.80g)/(216.59 g/mol) = 0.0083106 mole

Assuming the reaction can be completed, this amount of HgO can be lasted for:
0.0083106*96485*2/0.15e-3 = 1.069e7 (s) = 2969.82 (hr)

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