Instructions: Read the instructions carefully and answer the following questions in the space provided. Then attach the PDF file on Discussion Board for peer discussion and peer review.
In a few sentences define each of the three main classes of isomers below. Configurational Isomers can be further classified as either enantiomers or diastereomers and should also be discussed. Your descriptions should enable your peers to quickly and easily compare and contrast the differences between the different classes of isomers.
A. Constitutional Isomers (or Structural Isomers)
B. Conformational Isomers
C. Configurational Isomers (or Stereoisomers)
I. Enantiomers
II. Diastereomers

See explanation

Explanation:

A. Constitutional or structural isomers have the same molecular formula but different structural formulas.

B. Conformational isomers are compounds having the same atom to atom connectivity but differ by rotation about one or more single bonds.

C. Stereo isomers are compounds having the same molecular mass and atom to atom connectivity but different arrangement of atoms and groups in space.

I. Enantiomers are stereo isomers (optical isomers particularly) that are non-superimposable mirror images of each other.

II. Diasteromers are optical isomers that are not mirror images of each other.

Both diasteromers and enantiomers are types of optical isomers which in turn is one of the types of stereo isomers.

Stereo isomers differ from conformational isomers in that the arrangement of atoms in stereo isomers is permanent while conformational isomers results from free rotations in molecules about single bonds.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

beta decay is the answer