Given:

• The mass of H2 ⇒m=3.86×10−4 g

The number of moles of H2:

⇒n=(3.86×10−4 g)(2 g/mol)⇒n=1.93×10−4 mol

The reaction between the N2 and H2 is:

N2+3H2→2NH3

From the above reaction, the ratio of the moles of H2 and NH3 is:

⇒moles of H2moles of NH3=32

Now the 1.93×10−4 mol of H2 is taking part in the reaction, thus the produced moles of NH3 is computed as:

⇒moles of NH3=(23)(moles of H2)⇒moles of NH3=(23)(1.93×10−4 mol)⇒moles of NH3≈1.287×10−4 mol

Thus the number of molecules produced of NH3 is computed as:

⇒N=(1.287×10−4 mol)(6.022×1023 mol−1)⇒N≈7.75×1019 molecules