How many grams of nh3 can be produced from 3.72 mol of n2 and excess h2.

The problem is poorly written. The question in the title and #2 do not state which element that we need to calculate the quantity for. Since this is the case, I will only solve #3. 

3H₂(g) + N₂(g) --> 2NH₃(g). 

Since the molar mass of H₂ is 2 * 1.01 g/mol = 2.02 g/mol, we see that 7.52 x 10^(-4) g of H₂ corresponds to [7.52 x 10^(-4)]/2.02 = 3.72 x 10^(-4) mol of H₂. From the mole ratio, we see that 2/3 * 3.72 x 10^(-4) = 2.48 x 10^(-4) mol of NH₃ is produced. 

Finally, this corresponds to [2.48 x 10^(-4)] * (6.022 x 10^23) = 1.49 x 10^20 molecules of NH₃. 

I hope this helps!