Question

Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors. The reaction is given below.
Fe2O3(s) + 2 Al(s) 2 Fe(l) + Al2O3(s)

What masses of iron(III) oxide and aluminum must be used to produce 30.0 g iron?

aluminum
________g?

iron(III) oxide
________g?

What is the maximum mass of aluminum oxide that could be produced?
________g?

Answer 1

Answer:

a) 42.7g of iron(III) oxide ($Fe_{2}O_{3}$) must be used to produce 30.0 g iron

b) 14.4g of aluminum (Al) must be used to produce 30.0 g iron

c) 27.2g of aluminum oxide is the maximum mass of aluminum oxide that could be produced

Explanation:

1. First write the chemical equation for the thermite reaction, that is:

$Fe_{2}O_{3}_{(s)}+_{2}Al_{(s)}=_{2}Fe_{(l)}+Al_{2}O_{3}_{(s)}$

2. Then, use the stoichiometry of the reaction to calculate the masses of iron(III) oxide and aluminum used to produce 30.g of iron:

- For the iron(III) oxide:

$30.0gFe*\frac{1molFe}{56gFe}*\frac{1molFe_{2}O_{3}}{2molesFe}*\frac{159.7gFe_{2}O_{3}}{1molFe_{2}O_{3}}=42.7g$

- For the aluminum:

$30.0gFe*\frac{1molFe}{56gFe}*\frac{2molesAl}{2molesFe}*\frac{27gAl}{1molAl}=14.4gAl$

3. To calculate the maximum mass of aluminum oxide that could be produced, is needed to determine the limiting reagent between the iron(III) oxide and the aluminum.

To determine the limiting reagent, is needed to take the mass of each element and divide it between its molar mass, then divide this obtained quantity by the stoichiometric coefficient taken from the chemical reaction. The smallest number will be the limiting reagent and the calculations must be done with this quantity.

- For the iron(III) oxide:

$42.7gFe_{2}O_{3}*\frac{1molFe_{2}O_{3}}{159.7gFe_{2}O_{3}}=0.27molesFe_{2}O_{3}$

$\frac{0.27}{1}=0.27$

- For the aluminum:

$14.4gAl*\frac{1molAl}{27gAl}=0.53molesAl$

$\frac{0.53}{2}=0.27$

As the quantities are equal there are not limiting reagent.

4. Calculate the maximum mass of aluminum oxide that could be produced:

$14.4gAl*\frac{1molAl}{27gAl}*\frac{1molAl_{2}O_{3}}{2molesAl}*\frac{102gAl_{2}O_{3}}{1molAl_{2}O_{3}}=27.2g Al_{2}O_{3}$