Quantity of sodium dichromate(VI) calculation

For partial oxidation the equation is:

Na₂Cr₂O₇ + 3C₃H₇OH + 4H₂SO₄ ⇒ Na₂SO₄ + Cr₂(SO₄)₃ + 3C₂H₅CHO + 7H₂O

The density of propan-1-ol is 0.804 g cm^-3. This means that each cm³ of propan-1-ol has a mass of 0.804 g. So 1.5 cm³ has a mass of 0.804 × 1.5 = 1.2 g.

C₃H₇OH = 60, so we have 1.2/60 = 0.02 mols of propan-1-ol.

The above equation shows that we need a third as much sodium dichromate for oxidation

Moles of Na₂Cr₂O₇.2H₂O required = 0.02/3 = 0.0067 mol

Na₂Cr₂O₇.2H₂O = 298

Mass of sodium dichromate required = 0.0067 × 298 = 2.0 g

We have used a bit more than this (3 g), but the distillation process should ensure that insufficient time is allowed for complete oxidation.