Me! determine the number of atoms in 4.27 g gold.

This can be done with some simple dimensional analysis. First you convert your gram of gold into moles of gold There are 196.97 grams of gold for every one mole of gold so you divide 4.27 by 196.67 to get 0.0217 moles of gold. You then multiply the moles of gold by Avogadro's number, 6.02 X 10^23. Avogadro's number indicates the number of atoms in one mole. Therefore, there are 1.307 X 10^22 atoms in 4.27 grams of gold.  

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.

answer; i believe the correct answer is (solid gold);