Question

Borneol can also be oxidized to camphor using other oxidizing agents, such as sodium dichromate in acid. How do you do a balanced equation for this oxidation (Cr2O7 2- is reduced to Cr3+) and calculate the weight of Na2Cr2O7*2H2O should be needed (theoretically) to oxidize 0.50 g of borneol?

Answer 1

The balanced equation for the oxidation of borneol to camphor using sodium dichromate in acid is: 8 CH₂OH + 2 Cr₂O₇²¯ + 3 H₃O⁺ → 8 CHO + 4 Cr³⁺ + 11 H₂O. To calculate the weight of Na₂Cr₂O₇·2H₂O needed to oxidize 0.50 g of borneol, use stoichiometry.

Step-by-step explanation:

The balanced equation for the oxidation of borneol to camphor using sodium dichromate in acid is:

8 CH₂OH + 2 Cr₂O₇²¯ + 3 H₃O⁺ → 8 CHO + 4 Cr³⁺ + 11 H₂O

To calculate the weight of Na₂Cr₂O₇·2H₂O needed to oxidize 0.50 g of borneol, we need to use stoichiometry:

1 mol of borneol is equivalent to 2 mol of Na₂Cr₂O₇·2H₂O, so we can set up a ratio:

0.50 g borneol * (1 mol Na₂Cr₂O₇·2H₂O / 2 mol borneol) * (294.1 g Na₂Cr₂O₇·2H₂O / 1 mol Na₂Cr₂O₇·2H₂O) = 36.76 g Na₂Cr₂O₇·2H₂O

Answer 2

The mass of the sodium dichromate that we need is 0.95 g.

The oxidation of borneol to camphor using sodium dichromate involves a chemical reaction in which the alcohol functional group in borneol is oxidized to a ketone functional group in camphor. This type of oxidation reaction is common for converting secondary alcohols to ketones using strong oxidizing agents.

The balanced equation is shown in the image attached.

Number of moles of borneol = 0.50 g /154 g/mol

= 0.0032 moles

Since the reaction is 1:1

Mass of the sodium dichromate used = 0.0032 moles * 298 g/mol

= 0.95 g