Current of 250. a flows for 24.0 hours at an anode where the reaction occurring is as follows: mn2+(aq) + 2h2o(l) → mno2(s) + 4h+(aq) + 2e–what mass of mno2 is deposited at this anode?

From Faraday's 1st law of electrolysis,
Total electricity passed into system = Q = IT = 250 X 24 X 60 X 60
                                                                       = 2.16 X 10^7 C

We know that, 96500 C = 1 F
∴ 2.16 X 10^7 C =  223.8 F

Now, number of moles of MnO2  deposited = 223.8/2=111.9 

Finally, 1 mole of MnO2 ≡ 86.94 g
∴ 111.9 mole of MnO2 ≡  111.9 X 86.94 = 9728 g

Thus, mass of MnO2 that will be deposited at anode = 9728 g

It shows that the experiment is designed properly.