How many bromine atoms are present in 39.9 g of ch2br2

                  2.77 × 10²³ Br Atoms

                 Mass of CH₂Br₂  =  39.9 g

                 M.Mass of CH₂Br₂  =  173.83 g.mol⁻¹

Step 1: Calculate Moles of CH₂Br₂ as,

                 Moles  =  Mass ÷ M.Mass

Putting values,

                 Moles  =  39.9 g ÷ 173.83 g.mol⁻¹

                 Moles  =  0.23 mol

Step 2: Calculate number of CH₂Br₂ Molecules,

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of CH₂Br₂ Molecules can be written as,

         Moles  =  Number of CH₂Br₂ Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of CH₂Br₂ Molecules,

          Number of CH₂Br₂ Molecules  =  Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting value of moles,

          Number of CH₂Br₂ Molecules  =  0.23 mol × 6.022 × 10²³ Atoms.mol⁻¹

          Number of CH₂Br₂ Molecules  =  1.38 × 10²³ CH₂Br₂ Molecules

Step 3: Calculate Number of Br Atoms:

As,

                               1 Molecule of CH₂Br₂ contains  =  2 Atoms of Br

So,

            1.38 × 10²³  Molecules of CH₂Br₂ will contain  =  X Atoms of Br

Solving for X,

                      X  =  (1.38 × 10²³ CH₂Br₂ × 2 Br) ÷ 1 CH₂Br₂

                       X =  2.77 × 10²³ Br Atoms

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).