To determine the number of miles of calcium nitrate required, we need to calculate the amount of calcium nitrate in miles that would react with 4.55 moles of Chromium (III) sulfate.

The balanced chemical equation for the reaction between calcium nitrate and chromium (III) sulfate is:

3Ca(NO3)2 + Cr2(SO4)3 → 3CaSO4 + 2Cr(NO3)3

From the stoichiometry of the balanced equation, we can see that 3 moles of calcium nitrate react with 1 mole of chromium (III) sulfate. Therefore, 4.55 moles of chromium (III) sulfate would require:

(4.55 moles Cr2(SO4)3) × (3 moles Ca(NO3)2 / 1 mole Cr2(SO4)3) = 13.65 moles Ca(NO3)2

Now, we need to convert moles of calcium nitrate to miles. The molar mass of calcium nitrate is approximately 236 g/mol. So, 13.65 moles of calcium nitrate is equivalent to:

(13.65 moles Ca(NO3)2) × (236 g/mol) = 3224 g Ca(NO3)2

Assuming a density of 2.5 g/cm³ for calcium nitrate, we can calculate the volume of 3224 g of calcium nitrate:

Volume = Mass/Density = (3224 g) / (2.5 g/cm³) = 1289 cm³

Finally, we can convert the volume to miles by multiplying by the appropriate conversion factor:

1 mile = 160934.4 cm

Therefore, 1289 cm³ = (1289 cm³) × (1 mile / 160934.4 cm³) ≈ 0.008 miles

So, approximately 0.008 miles of calcium nitrate would be required to react with 4.55 moles of Chromium (III) sulfate.