Calculate the volume of oxygen you would need, at 1.00 atm and 298 K, to completely oxidize 51g of glucose.
To calculate the volume of oxygen required to completely oxidize glucose, we first need to determine the balanced chemical equation for the combustion of glucose. The balanced equation is as follows:
C6H12O6 + 6O2 → 6CO2 + 6H2O
From the balanced equation, we can see that 6 moles of oxygen are required to oxidize 1 mole of glucose.
Next, we need to determine the molar mass of glucose. Glucose (C6H12O6) has a molar mass of 180.16 g/mol.
Now, we can calculate the moles of glucose using the given mass of 51 g and the molar mass:
moles of glucose = mass of glucose / molar mass of glucose
= 51 g / 180.16 g/mol
= 0.283 mol
Since 1 mole of glucose requires 6 moles of oxygen, we can determine the moles of oxygen needed:
moles of oxygen = moles of glucose x coefficient ratio
= 0.283 mol x 6
= 1.70 mol
Finally, we can use the ideal gas equation to calculate the volume of oxygen. The ideal gas equation is given by:
PV = nRT
Where:
P = pressure (1.00 atm)
V = volume of gas (to be calculated)
n = moles of gas (1.70 mol)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (298 K)
Rearranging the equation to solve for V, we have:
V = (nRT) / P
= (1.70 mol x 0.0821 L·atm/mol·K x 298 K) / 1.00 atm
= 42.47 L
Therefore, the volume of oxygen required to completely oxidize 51 g of glucose at 1.00 atm and 298 K is approximately 42.47 L.
To calculate the volume of oxygen needed to completely oxidize 51g of glucose at 1.00 atm and 298 K, we need to use stoichiometry and the ideal gas law equation.
First, let's write the balanced chemical equation for the combustion of glucose:
C6H12O6 + 6O2 -> 6CO2 + 6H2O
From the equation, we can see that 1 mole of glucose reacts with 6 moles of oxygen to produce 6 moles of carbon dioxide and 6 moles of water.
Step 1: Calculate the number of moles of glucose.
To find the number of moles of glucose, we will divide the given mass by its molar mass.
Molar mass of glucose (C6H12O6) = 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 g/mol
Number of moles of glucose = mass of glucose / molar mass of glucose
= 51 g / 180.18 g/mol
= 0.283 mol
Step 2: Determine the number of moles of oxygen required.
Since the balanced equation tells us that 1 mole of glucose reacts with 6 moles of oxygen, the number of moles of oxygen required will be 6 times the number of moles of glucose.
Number of moles of oxygen = 6 × number of moles of glucose
= 6 × 0.283 mol
= 1.698 mol
Step 3: Calculate the volume of oxygen at 1.00 atm and 298 K using the ideal gas law.
The ideal gas law is given by:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in K)
Rearranging the equation, we get:
V = (nRT) / P
Now, plug in the values:
n = 1.698 mol
R = 0.0821 L·atm/(mol·K)
T = 298 K
P = 1.00 atm
V = (1.698 mol × 0.0821 L·atm/(mol·K) × 298 K) / 1.00 atm
= 42.53 L
Therefore, you would need approximately 42.53 liters of oxygen at 1.00 atm and 298 K to completely oxidize 51g of glucose.
Write and balance the equation.
C6H12O6 + 6O2 ==> 6CO2 + 6H2O
mols glucose = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols glucose to mols O2.
Now convert mols O2 to volume (L) using PV = nRT and the conditions listed.