How many molecules (not moles) of nh3 are produced from 6.89×10−4 g of h2?

First, let's find the moles of hydrogen used to produce the ammonia
      mol = mass ÷ molar mass
if the mass of hydrogen = 6.89 × 10⁻⁴ g
& the molar mass of hydrogen = 2 g/mol
      then moles of hydrogen = (6.89 × 10⁻⁴ g) ÷ 2 g/mol
                           moles of H₂ = 3.445 × 10⁻⁴  mol

Next, balanced eq'n for the production of ammonia from hydrogen gas and nitrogen gas:
                N₂   +   3H₂   →  2NH₃

Now, the mole ratio (ratio of the coefficients used to balance the equation) of H₂ : NH₃ is  3 : 2

∴ if the moles of H₂ = 3.445 × 10⁻⁴
then based on the mole ratio, the moles of NH₃  = [(3.445 × 10⁻⁴) ÷ 3] × 2
                                                                                 = (1.14833 × 10⁻⁴) × 2
                                                      ∴ mole of NH₃   = 2.2967 × 10⁻⁴
                                                                                                                                       Now, # of molecules =  moles × Avogadro's Constant
         
∴ # of molecules of NH₃ = (2.2967 × 10⁻⁴)  ×  (6.02 × 10²³)
                                          =  1.383 × 10²⁰ molecules
       
      Thus, the molecules of NH₃ produced is ≈ 1.38 × 10²⁰ molecules.