One kind of battery used in watches contains mercur(II) oxide. As current flows, the mercury oxide is reduced to mercury.
HgO(s) + H2O(l) + 2e^- ==> Hg(l) + 2OH^-(aq)
If 2.5x10^-5 amperes flows continuously 1095 days, what mass of Hg(l) is produced?
(2.5x10^-5)(1095days)(24hours/day)(3600seconds/hour) = (mass of Hg)(1 mol Hg/mw Hg)(2 mol e^-/1 mol Hg)
2365.2 = (x grams Hg)(1 mol Hg/200.59gHg)(2 mol e-/1 mol Hg)
= 23.58 grams
Check your work. It looks to me as if you have omitted the 96,485 factor. The 2365.2 looks ok.
To calculate the mass of Hg(l) produced in this reaction, we need to use the given current (2.5x10^-5 A), the time (1095 days), and the stoichiometry of the reaction.
First, convert the time from days to seconds:
1095 days * 24 hours/day * 3600 seconds/hour = 94,608,000 seconds.
Now let's set up the stoichiometry calculation using the unit cancelation method:
2.5x10^-5 A * 94,608,000 s = x grams Hg * (1 mol Hg / 200.59 g Hg) * (2 mol e^- / 1 mol Hg)
Calculate the left-hand side:
2.5x10^-5 A * 94,608,000 s = 2,365.2 C
Substitute this value into the equation and solve for x:
2,365.2 C = x g Hg * (1 mol / 200.59 g Hg) * (2 mol e^- / 1 mol Hg)
To isolate x, multiply the molar masses and divide both sides by the conversion factors:
x = 2,365.2 C * (1 mol Hg / 200.59 g Hg) * (1 mol e^- / 2 mol Hg)
x = 23.58 g Hg
Therefore, the mass of Hg(l) produced in this reaction is 23.58 grams.