How many molecules of f2 react with 66.6 g nh3 ? (stoichiometry)

5.89 × 10^23 molecules of F₂

The equation for the reaction between fluorine (F₂) and ammonia (NH₃) is given by;

5F₂ + 2NH₃ → N₂F₄ + 6 HF

We are given 66.6 g NH₃

We are required to determine the number of fluorine molecules

Moles = Mass ÷ Molar mass

Molar mass of ammonia = 17.031 g/mol

Moles of NH₃ = 66.6 g ÷ 17.031 g/mol

                      = 3.911 moles

From the equation 5 moles of Fluorine reacts with 2 moles of ammonia

Therefore,

Moles of fluorine = Moles of Ammonia × 5/2

                            = 3.911 moles × 5/2

                           = 9.778 moles

We know that 1 mole of a compound contains number of molecules equivalent to the Avogadro's number, 6.022 × 10^23 molecules

Therefore;

1 mole of F₂ = 6.022 × 10^23 molecules

Thus,

9.778 moles of F₂ = 9.778 moles × 6.022 × 10^23 molecules/mole

                              = 5.89 × 10^23 molecules

Therefore, the number of fluorine molecules needed is 5.89 × 10^23 molecules