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

Molarity is moles of solute over liters of solution. idk if that is what you are talking about, but you can convert moles of solute to grams of solute using dimensional analyzes by using molar mass found on the periodic table.

H tricky one let me think and get back to u