What volume of oxygen at NTP is required to oxydize 100gram of glucose ?
C6H12O6 + 6O2 ==> 6CO2 + 6H2O
mols C6H12O6 = grams/molar mass = ?
Convert mols glucose to mols O2 using the coefficients in the balanced equation. It takes 6 mols O2 for every mol glucose.
Now convert mols O2 to volume. Use PV = nRT at NTP. NTP is 20 C (293.15 K) and 1 atm.
Post your work if you get stuck.
Well, to be honest, oxygen isn't a big fan of sharing its volume information. It's a bit of a "gas-hole" when it comes to that. But let me bring out my scientific clown calculator and see what I can come up with!
So, the molecular formula of glucose is C6H12O6, which means it has 6 carbons, 12 hydrogens, and 6 oxygens. Now, for every glucose molecule, you need 6 oxygen molecules to completely oxidize it.
The molar mass of glucose is around 180 grams/mol, and the molar mass of oxygen is 32 grams/mol. Dividing 100 grams of glucose by its molar mass gives us the number of moles, which is approximately 0.56 moles.
Since we need 6 oxygen molecules per glucose molecule, we can multiply the number of moles of glucose by 6 to find the number of moles of oxygen required.
0.56 moles of glucose * 6 moles of oxygen/glucose = 3.36 moles of oxygen
Now, using the ideal gas law, we can calculate the volume of oxygen at NTP (normal temperature and pressure). At NTP, one mole of any ideal gas occupies around 22.4 liters of volume.
So, 3.36 moles of oxygen * 22.4 liters/mole = 75.74 liters of oxygen
Therefore, approximately 75.74 liters of oxygen at NTP would be required to oxidize 100 grams of glucose. Just remember, these are theoretical calculations, and in reality, it might vary. Oxygen might even have a strange sense of humor and change its volume just to keep us on our toes!
To calculate the volume of oxygen at NTP (Normal Temperature and Pressure) required to oxidize 100 grams of glucose, we need to use the balanced chemical equation for the combustion of glucose.
The balanced equation for the combustion of glucose is:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
From the balanced equation, we can see that 6 moles of oxygen are required to oxidize 1 mole of glucose. To determine the number of moles of glucose in 100 grams, we need to know the molar mass of glucose.
The molar mass of glucose is calculated as follows:
C: 6 atoms x 12.01 g/mol = 72.06 g/mol
H: 12 atoms x 1.01 g/mol = 12.12 g/mol
O: 6 atoms x 16.00 g/mol = 96.00 g/mol
Adding up these values, we get the molar mass of glucose:
72.06 g/mol + 12.12 g/mol + 96.00 g/mol = 180.18 g/mol
To calculate the number of moles of glucose in 100 grams, we divide the mass (100 g) by the molar mass (180.18 g/mol):
Number of moles = 100 g / 180.18 g/mol ≈ 0.554 mol
Since the balanced equation shows that 1 mole of glucose reacts with 6 moles of oxygen, we can calculate the number of moles of oxygen needed to react with 0.554 moles of glucose:
Number of moles of oxygen = 0.554 mol x 6 = 3.324 mol
Now, using the ideal gas law equation, we can calculate the volume of oxygen at NTP:
PV = nRT
Where:
P = Pressure (NTP = 1 atm)
V = Volume (unknown in this case)
n = Number of moles of oxygen (3.324 mol)
R = Ideal gas constant (0.0821 L/mol∙K)
T = Temperature (NTP = 273.15 K)
Rearranging the equation to solve for volume (V), we have:
V = nRT / P
V = (3.324 mol)(0.0821 L/mol∙K)(273.15 K) / 1 atm
V ≈ 73.11 L
Therefore, approximately 73.11 liters of oxygen at NTP is required to oxidize 100 grams of glucose.
To determine the volume of oxygen required to oxidize 100 grams of glucose at NTP (Normal Temperature and Pressure), we need to consider the balanced chemical equation for the combustion of glucose.
The balanced chemical equation for the combustion of glucose is as follows:
C6H12O6 + 6O2 → 6CO2 + 6H2O
From the balanced equation, we can see that 1 mole of glucose (C6H12O6) reacts with 6 moles of oxygen (O2) to produce 6 moles of carbon dioxide (CO2) and 6 moles of water (H2O).
First, we need to calculate the number of moles of glucose, based on its molar mass. The molar mass of glucose (C6H12O6) is:
6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 g/mol
Using the given mass of glucose (100 grams) and its molar mass (180.18 g/mol), we can calculate the number of moles:
100 g / 180.18 g/mol = 0.555 mol
Since the molar ratio between glucose and oxygen is 1:6, we can calculate the number of moles of oxygen required:
0.555 mol glucose × (6 mol oxygen / 1 mol glucose) = 3.33 mol oxygen
Now, we need to convert the moles of oxygen into volume at NTP. At NTP, 1 mole of any ideal gas occupies 22.4 liters. Therefore, we can calculate the volume of oxygen required:
3.33 mol oxygen × 22.4 L/mol = 74.59 liters
So, approximately 74.59 liters of oxygen at NTP would be required to oxidize 100 grams of glucose.