Question

Iron reacts slowly with oxygen and water to form a compound commonly called rust. For 45.2 kg of rust, Calculate

1. The moles of compound
2. The moles of${\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$
3. The grams of iron

Short answer

1. Moles of rust is$195.01\mathrm{mol}{\mathrm{Fe}}_2{\mathrm{O}}_3.4{\mathrm{H}}_2\mathrm{O}$
2. Moles of ${\mathrm{Fe}}_2{\mathrm{O}}_3=195.01\mathrm{mol}{\mathrm{Fe}}_2{\mathrm{O}}_3$
3. Mass of Fe$=2.18\times {10}^4\mathrm{g}$

Step-by-step solution

Definition of moles

A standard scientific unit for measuring large quantities of very small entities, such as atoms, molecules or other specified particles.

Calculations of moles of compound

The molar mass of rust is 231.78 g/mol

The moles of a compound is:

Moles of rust is:

$=4.52\times {10}^4\mathrm{g}{\mathrm{Fe}}_2{\mathrm{O}}_3.4{\mathrm{H}}_2\mathrm{O}\left(\frac{\mathrm{mol}{\mathrm{Fe}}_2{\mathrm{O}}_3.4{\mathrm{H}}_2\mathrm{O}}{231.78\mathrm{g}{\mathrm{Fe}}_2{\mathrm{O}}_3.4{\mathrm{H}}_2\mathrm{O}}\right)$

The moles of rust is$195.01\mathrm{mol}{\mathrm{Fe}}_2{\mathrm{O}}_3.4{\mathrm{H}}_2\mathrm{O}$

Calculations of moles of Fe2O3

Moles of is:

$=195.01\mathrm{mol}{\mathrm{Fe}}_2{\mathrm{O}}_3.4{\mathrm{H}}_2\mathrm{O}\left(\frac{\mathrm{mol}{\mathrm{Fe}}_2{\mathrm{O}}_3.4{\mathrm{H}}_2\mathrm{O}}{231.78\mathrm{g}{\mathrm{Fe}}_2{\mathrm{O}}_3.4{\mathrm{H}}_2\mathrm{O}}\right)$

Calculations of moles of Fe

Moles of Fe is:

$=195.01\mathrm{mol}{\mathrm{Fe}}_2{\mathrm{O}}_3.4{\mathrm{H}}_2\mathrm{O}\left(\frac{2\mathrm{mol}\mathrm{Fe}}{1\mathrm{mol}{\mathrm{Fe}}_2{\mathrm{O}}_3.4{\mathrm{H}}_2\mathrm{O}}\right)$

The moles of Fe$=390.02\mathrm{mol}$

Determination of the mass of iron

Multiply the calculated number of moles of ion by its molar mass which is$55.85\frac{\mathrm{g}}{\mathrm{mol}}$

The mass of Fe is:

$=390.02\mathrm{mol}\mathrm{Fe}\left(\frac{55.85\mathrm{g}\mathrm{Fe}}{\mathrm{mol}\mathrm{Fe}}\right)$

The mass of Fe$=2.18\times {10}^4\mathrm{g}$