Use the half-reactions of the reaction Upper A u (upper O upper H) subscript 3 plus upper H upper I right arrow upper A U plus upper I subscript 2 plus upper H subscript 2 upper O. to answer the questions. Reduction half-reaction:Upper A u superscript plus 3, plus 3 e superscript minus right arrow upper A u.
Oxidation half-reaction:2 upper i right arrow upper I subscript 2 plus 2 e superscript minus.

How many electrons does each gold atom gain?
How many electrons does each iodine atom lose?
What is the total number of electrons that are moved in the oxidation-reduction reaction?

Complete the final balanced equation based on the half-reactions.
_Au(OH)3+_HI Right arrow_ Au +_ I2 + _ H2O

How many electrons does each gold atom gain?

  3

How many electrons does each iodine atom lose?

   1

What is the total number of electrons that are moved in the oxidation-reduction reaction?

   6

Complete the final balanced equation based on the half-reactions.

2Au(OH)3+6 HI 2 Au +1 3I2 +6 H2O

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.

answeridentity;