How many molecules of F2 react with 66.6 g NH3? (stoichiometry)

5.89 * 10^23 molecules of F₂

Explanation:

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

Step 1: Moles of Ammonia

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

Step 2: Moles of Fluorine

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

Step 3: Number of molecules of fluorine

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