Question

Before arc welding was developed, a displacement reaction involving aluminum and iron (III) oxide was commonly used to produce molten iron (the thermite process). This reaction was used, for example, to connect sections of iron railroad track. Calculate the mass of molten iron produced when 1.50 kg of aluminum reacts with 25.0 mol of iron (III) oxide.

Short answer

You need to calculate the mass of molten iron produced when 1.50 kg of aluminum reacts with 25.0 mol of iron (III) oxide.

Step-by-step solution

Balanced equation

Aluminium reacts with iron oxide and forms aluminium oxide and iron. The balanced equation is

$2AI\left(s\right)+F{e}_2{O}_3\left(s\right)\to 2Fe\left(s\right)+A{I}_2{O}_3\left(s\right)$

Calculation of moles of Al

According to the question,

Mass of Al = 1.50kg = 1500g

Molecular mass of Al = 26.98g/mol

Again you know,

$moles=\frac{mass\left(given\right)}{mass\left(molar\right)}$

Moles of Al

$$\begin{gathered} =\frac{1500}{26.98}\left(mol\right) \\ =55.5967\left(mol\right) \end{gathered}$$

Hence, moles of Al are 55.5967mol.

Calculation of moles of Fe when Al is limiting reagent

According to the balanced equation

$2AI\left(s\right)+F{e}_2{O}_3\left(s\right)\to 2Fe\left(s\right)+A{I}_2{O}_3\left(s\right)$

2 moles of Al produces 2 moles Fe

Now, moles of Fe

$$\begin{gathered} =55.5967\times \frac{2}{2}\left(mol\right) \\ =55.5967\left(mol\right) \end{gathered}$$

Hence, moles of Fe when Al is limiting reagent are 55.5967mol.

Calculation of moles of Fe when Fe2O3 is limiting reagent

According to the balanced equation

$2AI\left(s\right)+F{e}_2{O}_3\left(s\right)\to 2Fe\left(s\right)+A{I}_2{O}_3\left(s\right)$

1 mole of Fe₂O₃ produces 2 moles Fe

According to the question;

Moles of Fe₂O₃ are 25moles.

Now, moles of Fe

$$\begin{gathered} =25\times \frac{2}{1}\left(mol\right) \\ =50\left(mol\right) \end{gathered}$$

Hence, moles of Fe when Fe₂O₃ is limiting reagent are 50mol.

Conclusion

Fe₂O₃ is the limiting reagent in this reaction as moles of Fe is less in terms of Fe₂O₃.

Calculation of mass of Fe

Moles of Fe = 50moles

Molecular mass of Fe = 55.85g/mol

Again you know,

Mass = moles x mass(molar)

Mass of Fe

$$\begin{gathered} =50\times 55.85\left(g\right) \\ =2792.5\left(g\right) \\ =2.79\left(kg\right) \end{gathered}$$

Hence, mass of Fe is 2.79kg.

Conclusion

Hence, the mass of molten iron produced when 1.50 kg of aluminum reacts with 25.0 mol of iron (III) oxide is 2.79kg.