Consider the following reaction: iron (s) + chlorine (g) à iron (iii) chloride a. write the balanced chemical equation. b. 25.0 g of iron reacts with excess chlorine gas. a. calculate the moles of iron reactant. b. calculate the moles of iron (iii) chloride. c. calculate the mass of iron (iii) chloride.

Explanation:

                  Fe + Cl₂ →  FeCl₃

a. The balanced equation

    All chemical equations obey the law of conservation of matter. To balance this above equation, we either inspect or use mathematical method to obtain a balanced equation.

We put coefficients a, b and c at the back of the compounds as shown below:

                    aFe + bCl₂ →  cFeCl₃

For Fe:

      a = c                          (i)

For Cl:

     2b = 3c                        (ii)

let a = 1, c= 1

Solving for the unknown b, we have:

          b =

           b =

The complete reaction equation is therefore:

            Fe + Cl₂ →  FeCl₃

                                     or

                       2Fe + 3Cl₂ →  2FeCl₃

Problem a:

Mass of reacting iron = 25g

unknown: number of moles of iron

Solution

To find the number of moles, we apply the mole concept using the expression below:

To find the molar mass of the reactant iron, we know atomic mass of iron is 56g.

The molar mass is therefore, 56gmol⁻¹

number of moles of the iron reactant  = 0.45mole

Problem b:

From the balanced equation of the reaction:

             2Fe + 3Cl₂ →  2FeCl₃

      2 moles of Fe produces 2 moles of FeCl₃

     So 0.45mole of Fe would produce 0.45 mole of FeCl₃

Problem C:

Applying the mole concept;

      mass of FeCl₃ = number of moles of FeCl₃ x molar mass of FeCl₃

     number of moles of FeCl₃ = 0.45mole

     molar mass of FeCl₃ = ?

Atomic mass of Fe = 56g

                          Cl = 35.5g

      Molar mass of FeCl₃ = 56 + (3 x 35.5) = 56 + 106.5 = 162.5gmol⁻¹

mass of FeCl₃ = 0.45 x 162.5 = 73.1g

the answer is b. dilution