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.

Relax

Respuesta :

Answer:

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 = [tex]\frac{3c}{2}[/tex]

           b = [tex]\frac{3}{2}[/tex]

The complete reaction equation is therefore:

            Fe + [tex]\frac{3}{2}[/tex]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:

[tex]number of moles = \frac{mass}{molar mass} \\[/tex]

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

The molar mass is therefore, 56gmol⁻¹

[tex]number of moles = \frac{25}{56} \\[/tex]

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