What is the identity of the element which had the following electron configuration? 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5

The element with electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 is manganese (25Mn).

Explanation:

Step 1: Data given

The element with electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5

has 25 electrons.

This element has 2 electrons on the first shell, 8 electrons on the second shell, 13 electrons on the third shell and 2 electrons on the outer shell (valence electrons).

This means this element is part of group VII.

The element with 25 electrons, we can find on the periodic table, with atomic number 25.

The element with electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 is manganese (25Mn).

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

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons