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Question: Identify the product and its quantity obtained when 9.2 g ethanol is treated with acidified potassiu...

Identify the product and its quantity obtained when 9.2 g ethanol is treated with acidified potassium dichromate under ideal conditions

A

Ethanal, 3.2 g

B

Ethanoic acid, 3.2 g

C

Ethanoic acid, 12.0 g

D

Ethanol, 4.6 g

Answer

Ethanoic acid, 12.0 g

Explanation

Solution

  1. Calculate moles of ethanol:

    Moles of ethanol=9.2 g46 g/mol=0.2 mol\text{Moles of ethanol} = \frac{9.2 \text{ g}}{46 \text{ g/mol}} = 0.2 \text{ mol}

  2. Oxidation Reaction:

    Ethanol (CH3CH2OH\text{CH}_3\text{CH}_2\text{OH}) is oxidized by acidified potassium dichromate to ethanoic acid (CH3COOH\text{CH}_3\text{COOH}) under ideal conditions:

    CH3CH2OH→CH3COOH\text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{COOH}

    The molar ratio is 1:1.

  3. Calculate mass of ethanoic acid:

    Molar mass of ethanoic acid = 60 g/mol

    Mass=0.2 mol×60 g/mol=12.0 g\text{Mass} = 0.2 \text{ mol} \times 60 \text{ g/mol} = 12.0 \text{ g}

0.2 mol ethanol yields 0.2 mol ethanoic acid which weighs 12.0 g.