Chapter 9, Problem 21A

Interpretation Introduction

Interpretation:

The amount of iron (III) chloride formed when 15.5 mg of iron reacts with excess chlorine gas needs to be deduced based on the given reaction.

Concept Introduction:

• A chemical reaction is expressed as a chemical equation having reactants and products on left and right side of the reaction arrow respectively.

  $\text{Reactants }\to \text{ Products}$

• The coefficient of a balanced chemical equation, i.e., the stoichiometry gives the number of reactants and products involved in the reaction.
• Chemical equations can therefore be used to determine the amount of products formed from a known quantity of reactants

Answer to Problem 21A

Mass of FeCl₃ = 44.9 mg i.e. 45.0 mg

Explanation of Solution

The given reaction is:

  ${\text{Fe(s) + Cl}}_{\text{2}}\text{(g) }\to {\text{FeCl}}_{\text{3}}\text{(s) }$

The balanced equation is:

  ${\text{2Fe(s) + 3Cl}}_{\text{2}}\text{(g) }\to 2{\text{FeCl}}_{\text{3}}\text{(s) }$

Here since Cl₂ is present in excess and Fe is the limiting reagent. The amount of FeCl₃ formed will be determined by the amount of Fe reacted.

Step 1: Calculate the moles of Fe present:

Mass of Fe present = 15.5mg = 0.0155 g

Atomic weight of Fe = 55.85 g/mole

  $\begin{array}{l}\text{Moles of Fe = }\frac{\text{Mass of Fe}}{\text{Molar Mass Fe}}\\ =\frac{0.0155\text{ g}}{55.85\text{ g/mol}}=0.000277\text{ moles}\end{array}$

Step 2: Calculate the moles of FeCl₃ formed:

Based on the reaction stoichiometry:

Fe and FeCl3 are present in a 1:1 molar ratio

Therefore, 0.000277 moles of Fe would produce 0.000277 moles of FeCl₃

Step 3: Calculate the mass of FeCl₃ formed:

Moles of FeCl₃ formed= 0.000277 moles

Molecular weight of FeCl3 = 162.2 g/mol

  $\begin{array}{l}\text{Mass of FeCl3 = Moles }\times \text{ Molecular weight}\\ =0.000277\text{ moles }\times \text{ 162}\text{.2 g/mol = 0}\text{.0449 g}\end{array}$

Conclusion

Therefore, mass of FeCl3 formed is around 44.9 mg.