A sample of 2.30 g of ethanol was mixed with an excess of aqueous acidified potassium dichromate (VI) and refluxed to give ethanoic acid. If the yield of product collected was 60.0%, what mass of product was collected?(Ethanol = 46, Ethanoic acid = 60)

Ethanol ($\text{CH}_3\text{CH}_2\text{OH}$) is oxidized to ethanoic acid ($\text{CH}_3\text{COOH}$) using an oxidizing agent like acidified potassium dichromate(VI). The balanced equation is:

$$\text{CH}_3\text{CH}_2\text{OH} + 2[\text{O}] \longrightarrow \text{CH}_3\text{COOH} + \text{H}_2\text{O}$$

From the equation, 1 mole of ethanol produces 1 mole of ethanoic acid.

The molar mass of ethanol ($\text{C}_2\text{H}_5\text{OH}$) is given as 46 g/mol.
The mass of ethanol is 2.30 g.

$$\text{Moles of ethanol} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{2.30 \text{ g}}{46 \text{ g/mol}} = 0.050 \text{ mol}$$

Since the mole ratio of ethanol to ethanoic acid is 1:1, the theoretical moles of ethanoic acid produced will be equal to the moles of ethanol.

$$\text{Theoretical moles of ethanoic acid} = 0.050 \text{ mol}$$

The molar mass of ethanoic acid ($\text{CH}_3\text{COOH}$) is given as 60 g/mol.

$$\text{Theoretical mass of ethanoic acid} = \text{Moles} \times \text{Molar mass} = 0.050 \text{ mol} \times 60 \text{ g/mol} = 3.0 \text{ g}$$

The yield of the product was 60.0%.

$$\text{Actual mass collected} = \text{Theoretical mass} \times \frac{\text{Yield}}{100\%} = 3.0 \text{ g} \times \frac{60.0}{100} = 1.8 \text{ g}$$

The mass of product collected was 1.8 g.

$$\text{1.8 g}$$