Chemistry - Mole Concept Question with Solution | TestHub

The balanced chemical equation is:

$${\mathrm{Fe}}_3{\mathrm{O}}_4(\mathrm{s})+4{\mathrm{CO}}_{(\mathrm{g})}\to 3{\mathrm{Fe}}_{(1)}+4{\mathrm{CO}}_{2(\mathrm{g})}$$

Given:

Molar mass of ${\mathrm{Fe}}_3{\mathrm{O}}_4=232\mathrm{g}{\mathrm{mol}}^{-1}$

Molar mass of $\mathrm{CO}=28\mathrm{g}{\mathrm{mol}}^{-1}$

Molar mass of $\mathrm{Fe}=56\mathrm{g}{\mathrm{mol}}^{-1}$

Initial amounts:

Mass of ${\mathrm{Fe}}_3{\mathrm{O}}_4=2.32\times 10^3\mathrm{kg}=2.32\times 10^6\mathrm{g}$

Mass of $\mathrm{CO}=2.8\times 10^2\mathrm{kg}=2.8\times 10^5\mathrm{g}$

1. Calculate moles of reactants:

Moles of ${\mathrm{Fe}}_3{\mathrm{O}}_4=\frac{2.32\times 10^6\mathrm{g}}{232\mathrm{g}{\mathrm{mol}}^{-1}}=1\times 10^4\mathrm{mol}$

Moles of $\mathrm{CO}=\frac{2.8\times 10^5\mathrm{g}}{28\mathrm{g}{\mathrm{mol}}^{-1}}=1\times 10^4\mathrm{mol}$

2. Determine the limiting reactant:

From the stoichiometry, 1 mole of ${\mathrm{Fe}}_3{\mathrm{O}}_4$ reacts with 4 moles of $\mathrm{CO}$.

For $1\times 10^4$ moles of ${\mathrm{Fe}}_3{\mathrm{O}}_4$, $4\times (1\times 10^4)=4\times 10^4$ moles of $\mathrm{CO}$ are required.

Since we only have $1\times 10^4$ moles of $\mathrm{CO}$, $\mathrm{CO}$ is the limiting reactant.

3. Calculate the mass of Fe produced based on the limiting reactant:

From the equation, 4 moles of $\mathrm{CO}$ produce 3 moles of $\mathrm{Fe}$.

So, $1\times 10^4$ moles of $\mathrm{CO}$ will produce $\frac{3}{4}\times (1\times 10^4)=0.75\times 10^4=7500\mathrm{mol}$ of $\mathrm{Fe}$.

Mass of $\mathrm{Fe}=molesofFe\times molarmassofFe$

Mass of $\mathrm{Fe}=7500\mathrm{mol}\times 56\mathrm{g}{\mathrm{mol}}^{-1}=420000\mathrm{g}$

Convert mass to kg:

Mass of $\mathrm{Fe}=420000\mathrm{g}=420\mathrm{kg}$

The value of 'x' is 420.