91Ó°ÊÓ

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

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

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.

According to the balanced equation

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

2 moles of Al produces 2 moles Fe

Now, moles of Fe

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

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

According to the balanced equation

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

1 mole of Fe₂O₃ produces 2 moles Fe

According to the question;

Moles of Fe₂O₃ are 25moles.

Now, moles of Fe

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

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

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

Moles of Fe = 50moles

Molecular mass of Fe = 55.85g/mol

Again you know,

Mass = moles x mass(molar)

Mass of Fe

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

Hence, mass of Fe is 2.79kg.

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.