The half life of iodine-125 is 60 days. What fraction of 125I nuclides would be left after 360 days?

THE FRACTION OF IODINE-125 REMAINING AFTER 360 DAYS IS 1/64.

Explanation:

half life of iodine- 125 = 60 days

Time elapsed for a decay = 360 days

using the formula:

                            Nt = No (1/2)^t/t1/2

The fraction left after a decay will be represented as:

Nt / No = (1/2) ^t/t1/2

Nt / No = (1/2) ^ 360 /60

Nt / No = (1/2) ^6

Nt / No = 1 / 64

The fraction of iodine-125 (Nt / No) remaining after 360 days is 1/ 64.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g