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 equation:

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

Using the molar volume of a gas at standard conditions (24.79 L mol^-1):

First convert 1006 mL to liters:

$1006 mL = 1.006 L$

Using the formula (n = \frac{V}{M_m} ):

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

From the balanced equation, we know that 1 mole of glucose produces 2 moles of ethanol. Thus, the amount of ethanol produced is:

$n_{C_2H_5OH} = 2 \times n_{CO_2} = 2 \times 0.040508087939 = 0.08101617588 moles$

Now to find the mass of ethanol produced, we use the molar mass of ethanol (C₂H₅OH), which is approximately 46.068 g mol^-1:

$m_{C_2H_5OH} = n_{C_2H_5OH} \times M_{C_2H_5OH} = 0.08101617588 moles \times 46.068 g mol^{-1} = 3.730388581 g$