A student conducted an experiment in the school laboratory under standard laboratory conditions (25°C, 100 kPa) to determine the volume of carbon dioxide gas produced during the fermentation of glucose - HSC - SSCE Chemistry - Question 25 - 2021 - Paper 1

Total volume of gas produced on Day 5 = 1006 mL = 1.006 L

Using the ideal gas law, we use the molar volume of gas at standard conditions (24.79 L mol⁻¹).

$n_{CO_2} = \frac{V_{CO_2}}{24.79 \, ext{L mol}^{-1}} = \frac{1.006 \, ext{L}}{24.79 \, ext{L mol}^{-1}} = 0.040508807939 \, \text{moles}$

From the balanced equation, the ratio of glucose to ethanol is 1:2. Therefore:

$n_{C_2H_5OH} = 2 \times n_{CO_2} = 2 \times 0.040508807939 = 0.08101761588 \, \text{moles}$

To find the mass of ethanol, we use its molar mass:

Molar mass of C₂H₅OH = 46.068 g mol⁻¹

Mass of ethanol:

$m_{C_2H_5OH} = n_{C_2H_5OH} \times \text{Molar mass} = 0.08101761588 \, \text{mol} \times 46.068 \, \text{g mol}^{-1} = 3.726390\, g$