When a rusty object was coated with phosphoric acid, all of the rust (taken as Fe2O3) on its surface was converted to iron phosphate according to the following balanced equation - Leaving Cert Chemistry - Question c - 2018

From the previous step, we found that 0.03 moles of iron were removed.

Next, we relate this to the moles of iron oxide (Fe2O3):

• The balanced equation shows that 2 moles of iron correspond to 1 mole of Fe2O3.

Thus, moles of iron oxide: $n(Fe2O3) = \frac{0.03 \, moles \ of \ iron}{2} = 0.015 \, moles \ of \ Fe2O3$

The molar mass of iron oxide (Fe2O3) = 160 g/mol. Now, we can calculate the mass of rust: $mass = n \times molar \ mass = 0.015 \, moles \times 160 \, g/mol = 2.4 \, g.$

To find the volume of phosphoric acid needed, we first determine the number of moles required:

• From the equation, 1 mole of Fe2O3 requires 2 moles of H3PO4. Thus, 0.015 moles of Fe2O3 would require: $n(H3PO4) = 2 \times 0.015 = 0.03 \, moles \ of \ H3PO4.$

Using the concentration of the phosphoric acid solution (6.0 M), we can find the volume:

• Molarity (M) = moles/volume (L)
• Rearranging gives: volume = moles / M

Therefore, $V = \frac{0.03 \, moles}{6.0 \, moles/L} = 0.005 \, L = 5.0 \, cm³.$

From the balanced equation, we see that for every mole of Fe2O3 that reacts, 3 moles of water are produced:

• From the previous calculations, we determined that 0.015 moles of Fe2O3 was used.

Thus, moles of water produced: $n(H2O) = 3 \times 0.015 = 0.045 \, moles.$

Now, using the density of water (1.0 g/cm³), we can find the volume:

• The mass of water produced: $mass = n \times molar \ mass = 0.045 \, moles \times 18 \, g/mol = 0.81 \, g.$

Finally, using density to find volume: $V = \frac{mass}{density} = \frac{0.81 \, g}{1.0 \, g/cm³} = 0.81 \, cm³.$