Suppose the half-life of an element is 10 years. how many half-lives will it take before only about 6% of the original sample remains? 3 half-lives 4 half-lives 5 half-lives 6 half-lives

We can solve this problem by using the equation as shown: A=Aoe^-(ln2/t)T where A is the final amount (0.06 gram), Ao is the initial amount (1g *assumed), t is the half-life (10 years ) and lastly T time elapsed (required). Plug in the values in the equation and you should get  40.5 years and this is approximately 4 half-lives. 

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explanation:

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