Chapter 4, Problem 17PS

In the thermite reaction, iron(III) oxide is reduced by aluminum to give molten iron

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

If you begin with 10.0 g of Fe₂O₃ and 20.0 g of AI,

1. (a) Which reactant is limiting?
2. (b) What mass of Fe can be produced?
3. (c) What mass of the excess reactant remains after the limiting reactant is consumed?
4. (d) Set up an amounts table for this problem.

Interpretation Introduction

 Interpretation:

 The limiting reactant of the reaction has to be identified.

 Concept introduction:

 Limiting reagent: limiting reagent is a reactant, which consumes completely in the chemical reaction.  The quantity of the product depends on this limiting reagent, it can be determined with the help of balanced chemical equation for the reaction.

Answer to Problem 17PS

 ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$is the limiting reactant in the given reaction.

Explanation of Solution

 Balanced chemical equation for the given reaction is,

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

 To find the mass of product formed in the reaction, the amount of each reactant involved in the given reaction should be determined.

 Therefore,

$\begin{array}{l}\text{Amount}\text{of}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{=}\text{10}\text{.0g}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{×}\frac{\text{1}\text{mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{\text{159}\text{.69}\text{g}}\text{=}\text{0}\text{.06262}\text{mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\\ \text{Amount}\text{of}\text{Al}\text{=}\text{20}\text{.0g}\text{Al}\text{×}\frac{\text{1}\text{mol}\text{Al}}{\text{26}\text{.98}\text{g}}\text{=}\text{0}\text{.74128}\text{mol}\text{Al}\end{array}$

The mass of $\text{Fe}$and ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ expected based on each reactant is,

$\begin{array}{l}\text{0}\text{.06262}\text{mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{×}\frac{\text{2}\text{mol}\text{Fe}\text{formed}}{\text{1}\text{mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{available}}\text{×}\frac{\text{55}\text{.8450}\text{g}\text{Fe}}{\text{1}\text{mol}\text{Fe}}\text{=}\text{6}\text{.994g}\text{Fe}\\ \\ \text{0}\text{.74128}\text{mol}\text{Al}\text{×}\frac{\text{1}\text{mol}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{formed}}{\text{2}\text{mol}\text{Al}\text{available}}\text{×}\frac{\text{101}\text{.96}\text{g}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}}{\text{1}\text{mol}{\text{AlCl}}_{\text{3}}}\text{=}\text{37}\text{.79g}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\end{array}$

 The amount of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$available produces fewer products than the amount of $\text{Al}$ available.

 Therefore, the reactant ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$is the limiting reactant.

Interpretation Introduction

 Interpretation:

 Mass of Fe produced in the reaction has to be determined.

 Concept Introduction:

• The number of moles of any substance can be determined  using the equation,

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

• 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 written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.

Answer to Problem 17PS

 The maximum mass of $\text{Fe}$that can be produced is$\text{6}\text{.994g}$.

Explanation of Solution

 From the previous calculation, the amount of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$available produces fewer products than the amount of $\text{Al}$ available..

The mass of $\text{Fe}$ expected based on each reactant is,

 The maximum mass of $\text{Fe}$that can be produced is$\text{6}\text{.994g}$.

Interpretation Introduction

 Interpretation:

 Mass of excess reactant remains after the consumption of limiting reactant has to be calculated.

Answer to Problem 17PS

 The excess mass of the reactant remains after the limiting reactant is consumed is $\text{16}\text{.6211}\text{g}\text{Al}\text{.}$

Explanation of Solution

 The mass of Al remains when all the ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ has been converted to product can be determined as follows,

  $\text{0}\text{.06262}\text{mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{×}\frac{\text{2}\text{mol}\text{Al}}{\text{1mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\text{×}\frac{\text{26}\text{.98g}\text{Al}}{\text{1}\text{mol}\text{Al}}\text{=}\text{3}\text{.3789}\text{g}\text{Al}\text{required}$

 Reaction began with $\text{20}\text{.0}\text{g}$of $\text{Al}$but only $3.3789\text{g}\text{Al}$is required by the limiting reactant, thus the mass of $\text{Al}$ that is present in excess is,

  $\text{20}\text{.0}\text{g}\text{Al}\text{present}\text{-}\text{3}\text{.3789}\text{g}\text{Al }\text{required}\text{=}\text{16}\text{.6211}\text{g}\text{Al}\text{left}$

Interpretation Introduction

 Interpretation:

 Amount table for the given problem has to be set up.

Explanation of Solution

The results of the above calculations can be summarized in an amounts table as follows,

Equation       ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\text{(s)}}\text{+}{\text{2Al}}_{\text{(s)}}\stackrel{}{\to }{\text{2Fe}}_{\text{(l)}}\text{+}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\text{(s)}}$
Initial amount (mol)	$0.06262$	$\text{0}\text{.7412}$	0	0
Change in amount upon reaction (mol)	$\text{-}0.06262$	$\text{-0}\text{.1252}$	$+0.1252$	$+0.06262$
Amount after complete reaction (mol)	0	$0.6160$	$0.1252$	$0.06262$