Ethanol ( C2H5OH) undergoes a reaction with oxygen gas in the air. Calculate the volume (l) of oxygen gas at 45 degrees celcius and mmHg that is required to react with 185 g ethanol. Then calculate the volume of air required to react with 185 g assuming that air is 21% by volume.
To calculate the volume of oxygen gas required to react with 185 g of ethanol, we need to determine the balanced equation for the reaction between ethanol and oxygen.
The balanced equation for the combustion of ethanol can be written as follows:
C2H5OH + 3O2 → 2CO2 + 3H2O
From the balanced equation, we can see that the stoichiometric ratio between ethanol and oxygen is 1:3. This means that one mole of ethanol reacts with three moles of oxygen.
To calculate the moles of ethanol, we can use the molar mass of ethanol, which is 46.07 g/mol.
Moles of ethanol = Mass of ethanol / Molar mass of ethanol
Moles of ethanol = 185 g / 46.07 g/mol = 4 moles
From the stoichiometry, we know that one mole of ethanol reacts with three moles of oxygen. Therefore, the moles of oxygen required can be calculated as follows:
Moles of oxygen = Moles of ethanol × Stoichiometric ratio
Moles of oxygen = 4 moles × 3 = 12 moles
Now, we can calculate the volume of oxygen gas at 45 degrees Celsius and mmHg using the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in mmHg)
V = Volume (in liters)
n = Moles of gas
R = Gas constant (0.0821 L·atm/mol·K)
T = Temperature (in Kelvin)
Before we can use the ideal gas law, we need to convert the temperature from Celsius to Kelvin:
Temperature in Kelvin = Temperature in Celsius + 273
Temperature in Kelvin = 45 + 273 = 318 K
Now, we can rearrange the ideal gas law equation to solve for volume:
V = (nRT) / P
Substituting the values into the equation:
V = (12 moles × 0.0821 L·atm/mol·K × 318 K) / P
Let's assume the pressure, P, is 760 mmHg:
V = (12 moles × 0.0821 L·atm/mol·K × 318 K) / 760 mmHg
V ≈ 65.51 L
So, the volume of oxygen gas required to react with 185 g of ethanol at 45 degrees Celsius and 760 mmHg is approximately 65.51 liters.
Now, let's calculate the volume of air required to react with 185 g of ethanol, considering that air is 21% by volume.
Since air is approximately 21% oxygen, we can calculate the volume of air required by dividing the volume of oxygen by the volume fraction of oxygen in air.
Volume of air = Volume of oxygen / Volume fraction of oxygen in air
Volume of air = 65.51 L / 0.21
Volume of air ≈ 312.43 L
Therefore, the volume of air required to react with 185 g of ethanol, assuming air is 21% by volume, is approximately 312.43 liters.