The decomposition of nitrogen dioxide to nitrogen monoxide and oxygen gas is a second order process as suspected in the previous question. one would expect the second half-life of this reaction to be the first half-life of this reaction.

Explanation:

For the given reaction

Now, expression for half-life of a second order reaction is as follows.

                      ....... (1)

Second half life of this reaction will be . So, expression for this will be as follows.

          =  ...(2)

where is the final concentration that is, here and is the initial concentration.

Hence, putting these values into equation (2) formula as follows.

        =

                      =     ...... (3)

Now, dividing equation (3) by equation (1) as follows.

            =                  

                                    = 3

or,                       = 3    

Thus, we can conclude that one would expect the second half-life of this reaction to be three times the first half-life of this reaction.

the answer is d : ) have a good da

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