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

The fermentation of glucose can be represented by the following equation:

$C_6H_{12}O_6(aq) \rightarrow 2C_2H_5OH(aq) + 2CO_2(g)$

First, we convert the volume of gas from mL to L:

$1006 \, \text{mL} = 1.006 \, \text{L}$

Now, use the molar volume of an ideal gas at standard conditions (24.79 L mol⁻¹):

$n_{CO_2} = \frac{1.006 \, \text{L}}{24.79 \, \text{L mol}^{-1}} = 0.040508087939 \, \text{moles of } CO_2$

Using the stoichiometry of the reaction, 2 moles of ethanol ($C_2H_5OH$) are produced for every mole of carbon dioxide. Thus,

$n_{C_2H_5OH} = \frac{n_{CO_2}}{2} = \frac{0.040508087939}{2} = 0.0202540439695 \, \text{moles of } C_2H_5OH$

The molar mass of ethanol ($C_2H_5OH$) is approximately 46.068 g mol⁻¹:

$m_{C_2H_5OH} = n_{C_2H_5OH} \times MM_{C_2H_5OH} = 0.0202540439695 \, \text{moles} \times 46.068 \, \text{g mol}^{-1} = 0.9322943471775 \, g$

Thus, the mass of ethanol produced is approximately 0.932 g.