One kind of battery used in watches contains mercury(II) oxide. As current flows, the mercury oxide is reduced to mercury.

Question

HgO(s) + H2O)(l) + 2e^- ==> Hg(l) + 2OH^-(aq)

If 2.5x10^-5 ameres flows continuously 1095 days, what mass of Hg(l) is produced?

(2.5x10^-5)(1095 days)(24hours/day)(3600seconds/hour) = (xg Hg)(1 mol Hg/200.59g Hg)(2mol e^-/1 mol Hg)(96,000)

2365.2 = (x)(957.18) = .4046g Hg

Answer 1

I would have used 96,485 but I think your math is still off. I'm getting something like 2.5 g.
It appears that you are dividing the wrong way.
2365.2 = x(957.18)
Divide both sides by 957.18.
(2365.2/957.18)=x(957.18/957.18)
2.47 = x (You divided 957.18/2365.2) = 0.4046)
Again, if you intend to use digits like 2365.2 and 957.18, I would also use 96,485 instead of rounding that to 96,000.