Q3.25P - Chemistry: Molecular Nature Of Matter And Change

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

Step-by-Step Solution

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 1: Definition of moles

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

Step 2: 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 $195.01 \mathrm{mol} {\mathrm{Fe}}_2{\mathrm{O}}_3.4{\mathrm{H}}_2\mathrm{O}$

Step 3: Calculations of moles of Fe 2 O 3

Moles of is:

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

Step 4: 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$=390.02\mathrm{mol}$

Step 2: 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$=2.18\times {10}^4\mathrm{g}$