What is the equilibrium constant expression for the ksp of sr3(po4)2?

The equilibrium constant expression for KSP of Sr3(PO4)2 is

KSP={(Sr^2+)^3 (PO4^3-)^2/ Sr3(PO4)2}

Explanation

write the ionic equation for Sr3(PO4)2

Sr3(PO4)2 → 3Sr^2+ + 2 PO4^3-

KSP is given by (concentration of the products raised to their coefficient /concentration of reactants raised to their coefficient)

The equilibrium constant expression for KSP of Sr3(PO4)2 is

KSP={(Sr^2+)^3 (PO4^3-)^2/ Sr3(PO4)2}

Explanation

write the ionic equation for Sr3(PO4)2

Sr3(PO4)2 → 3Sr^2+ + 2 PO4^3-

KSP is given by (concentration of the products raised to their coefficient /concentration of reactants raised to their coefficient)

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.

the answer is c

tell me if wrong