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