Question

consider the following equation.
fe₂o₃(s) + al(s) → fe(l) + al₂o₃(s)
calculate the quantity, in moles, of fe₂o₃ that will be required to produce 445 grams of fe.

Explanation:

Step1: Calculate moles of Fe

The molar mass of Fe is approximately 55.85 g/mol. Using the formula $n=\frac{m}{M}$, where $n$ is the number of moles, $m$ is the mass, and $M$ is the molar mass. For Fe, $m = 445$ g and $M=55.85$ g/mol. So, $n_{Fe}=\frac{445}{55.85}\approx 7.97$ mol.

Step2: Balance the chemical equation

The balanced chemical equation is $Fe_2O_3(s)+2Al(s)
ightarrow 2Fe(l)+Al_2O_3(s)$. From the balanced equation, the mole - ratio of $Fe_2O_3$ to $Fe$ is 1:2.

Step3: Calculate moles of $Fe_2O_3$

Let the moles of $Fe_2O_3$ be $n_{Fe_2O_3}$. Using the mole - ratio $\frac{n_{Fe_2O_3}}{n_{Fe}}=\frac{1}{2}$. So, $n_{Fe_2O_3}=\frac{n_{Fe}}{2}$. Substituting $n_{Fe}\approx 7.97$ mol, we get $n_{Fe_2O_3}=\frac{7.97}{2}=3.985$ mol.

Answer:

3.985 mol