The acid-dissociation constant for benzoic acid (c6h5cooh) is 6.3×10−5. calculate the equilibrium concentration of c6h5cooh in the solution if the initial concentration of c6h5cooh is 6.3×10−2 m .

Answer is: the equilibrium concentration of benzoic acid is 0,06294 M.

Chemical reaction: C₆H₅COOH(aq) ⇄ H⁺(aq) + C₆H₅COO⁻(aq).

Ka(C₆H₅COOH) = 6,3·10⁻⁵.

c(C₆H₅COOH) = 6,3·10⁻² M.

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.