Final answer:
To calculate the volume of carbon dioxide produced from fermenting 0.9g of glucose, the mass of glucose is first converted to moles, then using the stoichiometry of the balanced equation and the molar volume at r.t.p., the volume of CO2 is determined to be approximately 0.224 liters.
Step-by-step explanation:
To calculate the volume of carbon dioxide produced during the fermentation of glucose at room temperature and pressure (r.t.p.), we must first convert the mass of glucose to moles using its molar mass. The balanced equation for fermentation is:
C6H12O6 (aq) → 2C2H5OH (aq) + 2CO2 (g)
From the equation, 1 mole of glucose produces 2 moles of carbon dioxide. The molar mass of glucose (C6H12O6) is 180.156 g/mol. Thus, 0.9 g of glucose is:
0.9 g × (1 mol / 180.156 g) = 0.005 moles of glucose
Since 1 mole of glucose produces 2 moles of CO2, 0.005 moles of glucose will produce:
0.005 moles × 2 = 0.01 moles of CO2
At r.t.p., 1 mole of gas occupies 22.414 liters. Therefore, the volume of CO2 produced is:
0.01 moles × 22.414 L/mol = 0.22414 liters
So, 0.9 g of glucose ferments to produce approximately 0.224 liters of carbon dioxide at r.t.p.