Chapter 4, Problem 1PS

The reaction of iron(III) oxide with aluminum to give molten iron is known as the thermite reaction (page 172).

Fe₂O₃(s) + 2 Al(s) → 2 Fe(l) + Al₂O₃(s)

What amount of Al, in moles, is needed for complete reaction with 3.0 mol of Fe₂O₃? What mass of Fe, in grams, can be produced?

Interpretation Introduction

Interpretation:

From the given reaction the amount of Aluminum in moles needed for complete the reaction with $\text{3}\text{.0}\text{mol}\text{of}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ and the mass of $\text{Fe}$ produced in grams should be determined.

Concept introduction:

• $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$
• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• For chemical reaction balanced chemical reaction equation established in accordance with the Law of conservation of mass.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.

Answer to Problem 1PS

$\text{6}\text{.0}$ Mole aluminum is needed for complete the reaction with $\text{3}\text{.0}\text{mol}\text{of}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ $\text{335}\text{.04}\text{g}$ of $Fe$ produced in the given reaction.

Explanation of Solution

Balanced chemical equation for the given reaction is,

  ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\text{(s)}}\text{+}{\text{2Al}}_{\text{(s)}}\to {\text{2Fe}}_{\text{(l)}}\text{+}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\text{(s)}}$

Using the stoichiometric factor calculating the amount of Aluminum required based on the amount of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ as follows,

The amount of Aluminum in moles needed for complete the reaction with $\text{3}\text{.0}\text{mol}\text{of}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$

$\text{3}\text{.0}\text{mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{×}\frac{\text{2}\text{mol}\text{Al}}{\text{1}\text{mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\text{=}\text{6}\text{mol}\text{Al}$

The amount of $\text{Fe}$ in gram produced in the given reaction determined as follows,

  $\text{3}\text{.0}{\text{molFe}}_{\text{2}}{\text{O}}_{\text{3}}\text{}\text{×}\frac{\text{2}\text{mol}\text{Fe}}{\text{1}\text{mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\text{×}\frac{\text{55}\text{.84}\text{g}\text{Fe}}{\text{1}\text{mol}\text{Fe}}\text{=}\text{335}\text{.04}\text{g}\text{Fe}$

Therefore amount of $Fe$ in gram produced in the given reaction is $\text{335}\text{.04}\text{g}$.

Conclusion

The amount of Aluminum in moles needed for complete the reaction with and the mass of Fe produced in gram was determined.