A student has been asked to produce 185 mL of ethanol (MM = 46.068 g mol^-1) by fermenting glucose using yeast, as shown in the equation - HSC - SSCE Chemistry - Question 27 - 2023 - Paper 1

Using the molar mass:

$$n(ethanol) = \frac{m}{MM} = \frac{146 \, \text{g}}{46.068 \, \text{g mol}^{-1}} = 3.17 \, \text{mol}$$

From the reaction, 1 mole of glucose produces 2 moles of CO2. Therefore, 3.17 moles of ethanol would produce:

$$\text{moles of } CO_2 = 2 \times n(ethanol) = 2 \times 3.17 = 6.34 \, \text{mol}$$

Using the Ideal Gas Law to calculate the volume of CO2:

$$V = \frac{nRT}{P}$$

Where:

• R = 8.314 J K^-1 mol^-1
• T = 310 K
• P = 100 kPa = 100,000 Pa

Substituting the values:

$$V = \frac{6.34 \, \text{mol} \times 8.314 \, \text{JK}^{-1} \text{mol}^{-1} \times 310 \, ext{K}}{100 \, \text{kPa}} = 81.7 \, ext{L}$$