Question

For the oxidation of glucose (C6H12O6 + 6O2→6H2O + 6CO2), how many liters of carbon dioxide gas(CO2) will be produced when 950 g of glucose(C6H12O6) is oxidized completely?

Answer 1

Step-by-step explanation:

Write the balanced equation

C6H12O6 → 6CO2 + 6H2O

1 mol C6H12O6 produces 6 mol CO2

Molar mass C6H12O6 = 180 g/mol

mol CO2 in 950 g = 950 g / 180 g/mol = 5.277 mol

This will produce 5.277*6 = 31.67 mol CO2

Molar mass CO2 = 44 g/mol

Mass CO2 produced = 31.67 mol * 44 g/mol = 1393.33 g CO2 produced

Answer 2

When 950 g of glucose is oxidized completely, approximately 710.18 liters of carbon dioxide gas (CO2) will be produced.

Step-by-step explanation:

In the oxidation of glucose (C6H12O6 + 6O2→6H2O + 6CO2), 1 mole of glucose reacts with 6 moles of O2 to produce 6 moles of CO2. In order to determine how many liters of CO2 will be produced when 950 g of glucose is oxidized completely, we need to convert the mass of glucose to moles and then use the mole ratio to calculate the volume of CO2.

First, we need to find the number of moles of glucose by dividing the mass of glucose by its molar mass (180.16 g/mol):

Number of moles of glucose = 950 g / 180.16 g/mol = 5.277 moles

Now, we can use the mole ratio from the balanced equation to determine the number of moles of CO2:

Number of moles of CO2 = 5.277 moles * 6 moles CO2 / 1 mole C6H12O6 = 31.662 moles CO2

Finally, we can use the density of CO2 (1.96 g/L) to calculate the volume of CO2:

Volume of CO2 = 31.662 moles CO2 * (44.01 g/mol / 1.96 g/L) = 710.18 L