Determine the number of grams in 4.87 x 10^22 atoms of barium hydroxide Ba(OH)2' (fence post model)

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Explanation:

The  concentration   of  hydroxide ion  (-OH)  in  1.37 x10^-5 m solution  of Ba(OH)2  is  calculated  as   follows

write   the  equation  for  dissociation  of Ba(OH)2

Ba(OH)2 =  Ba^2+(aq)   + 2OH^-(aq)

by use  of  mole  ratio between Ba(OH)2 to OH^-  which   is  1:2  the   concentration  of OH^-  =  2x  ( 1.37 x10^-5) = 2.74  x10^-5 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.

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