Question

A digital watch draws 0.230 mA of current provided by a battery whose net reaction is:

HgO(s) + Zn(s) ----> ZnO(s) + Hg(l)

If a partially used battery contains 0.350 gm each of the four substances for how many more hours will the watch run? Hint: First determine the limiting reagent.

Answer 1

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

Molar mass of HgO is = 200.6 + 16 = 216.6 g/mol

Molar mass of Zn is = 65.4 g/mol

According to the balanced equation ,

1 mole of HgO reacts with 1 mole of Zn

                         OR

216.6 g of HgO reacts with 65.4 g of Zn

0.350 g of HgO reacts with Z g of Zn

So 0.350 - 0.105 = 0.245 g of Zn left unreacted , it is the excess reactant.So HgO is the limiting reactant.

Calculation of time taken :-

We know that

Where

           W = mass of Zn consumed = 1.105 g

           E = Equivalent weight of Zn = 65.4 / 2 = 32.7

           C = current = 0.230 mA = 0.230x10 ^-3 A

           t = time taken = ?

Plug the values we get