The al2o3 crystal structure (corundum) consists of an hcp arrangement of o2- ions; the al3+ ions occupy octahedral positions. what fraction of the available octahedral positions are filled with al3+ ions? in corundum, the al3+ ions do not fill the tetrahedral positions.

The fraction of the available octahedral positions are filled with ions is

Explanation:

If there are n numbers of spheres in HCP in packing:

Number of octahedral void = n

Number of tetrahedral void = 2n

Total number of voids = n + 2n = 3n

The crystal structure consists of an HCP arrangement of ions.

Number of spheres in HCP = 6

Number of = 6

Number of octahedral voids = 6

Number of tetrahedral void = 12

Number of aluminum ions when there are 6 ion =

Number of voids in which ions are present = 4

The fraction of the available octahedral positions are filled with ions =

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.