How many molecules (not moles) of nh3 are produced from 5.81×10−4 g of h2? express your answer numerically as the number of molecules?

The reaction of formation of NH₃ is as follows;
N₂ + 3H₂ ---> 2NH₃
the stoichiometry of H₂ to NH₃ is 3:2
for every 3 moles of H₂ that reacts, 2 moles of NH₃ are produced 
the mass of H₂ that has been used up = 5.81 x 10⁻⁴ g 
molar mass of H₂ = 2 g/mol
number of H₂ moles that were used up = 5.81 x 10⁻⁴ g/ 2 g/mol
                                                              = 2.905 x 10⁻⁴ mol
for every 3 mol of H₂ used up - 2 moles of NH₃
for every 1 mol of H₂ used up - 2/3 moles of NH₃
therefore for 2.905 x 10⁻⁴ mol of H₂ = 2/3 x 2.905 x 10⁻⁴ mol 
number of NH₃ moles produced = 1.94 x 10⁻⁴ mol
number of NH₃ molecules in 1 mole = 6.022 x 10²³ molecules 
therefore in 1.94 x 10⁻⁴ mol  = 6.022 x 10²³ molecules/mole x 1.94 x 10⁻⁴ mol

therefore number of NH₃ molecules produced = 1.168 x 10²⁰ NH₃ molecules  

Answer) by studying proton collisions hope this : )