Ethanol (C2H6O) is combusted with oxygen to produce carbon dioxide and wateraccording to the following reaction:
C2H6O + 3O2 -----> 2H2O + 3CO2
If 50g of ethanol is combusted, determine the volume of carbon dioxide that will be produced if the temperature is 25oC and the pressure is 100kPa.
1. mols ethanol = grams/molar mass
2. Use the coefficients in the balanced equation to convert mols ethanol to mols CO2.
3. Use PV = nRT and substitute n = mols and the other conditions and solve for volume in L. Don't forget T must be in kelvin.
To determine the volume of carbon dioxide produced when 50g of ethanol is combusted, we need to use the ideal gas law.
The stoichiometry of the balanced equation tells us that for every 1 mole of ethanol combusted, 3 moles of carbon dioxide are produced. To find the number of moles of ethanol, we need to calculate the molar mass of ethanol.
The molar mass of C2H6O = (2 x atomic mass of carbon) + (6 x atomic mass of hydrogen) + (1 x atomic mass of oxygen)
= (2 x 12.01 g/mol) + (6 x 1.01 g/mol) + (1 x 16.00 g/mol)
= 46.07 g/mol
Now, we can calculate the number of moles of ethanol:
Moles of ethanol = Mass of ethanol / Molar mass of ethanol
Moles of ethanol = 50 g / 46.07 g/mol
Moles of ethanol ≈ 1.086 mol (rounded to three decimal places)
Using the stoichiometry of the balanced equation, we know that 1 mole of ethanol will produce 3 moles of carbon dioxide. Therefore, 1.086 moles of ethanol will produce (1.086 mol ethanol x 3 mol CO2/mol ethanol) = 3.258 moles of carbon dioxide.
Now, we can use the ideal gas law to calculate the volume of carbon dioxide:
PV = nRT
Where:
P = pressure (in atm) = 100 kPa/101.325 kPa/atm ≈ 0.9869 atm (rounded to four decimal places)
V = volume (in liters) (the quantity we are trying to find)
n = number of moles = 3.258 moles
R = gas constant = 0.0821 L·atm/(mol·K) (at 25oC)
T = temperature (in Kelvin) = 25oC + 273.15 K = 298.15 K
Plugging in the values into the ideal gas law equation:
(0.9869 atm)·V = (3.258 mol)·(0.0821 L·atm/(mol·K))·(298.15 K)
Simplifying:
V ≈ (3.258 mol)·(0.0821 L·atm/(mol·K))·(298.15 K) / (0.9869 atm)
V ≈ 80.54 L (rounded to two decimal places)
Therefore, the volume of carbon dioxide produced when 50g of ethanol is combusted at 25oC and 100 kPa pressure is approximately 80.54 liters.
To determine the volume of carbon dioxide produced when 50g of ethanol is combusted, we need to follow these steps:
Step 1: Convert the mass of ethanol (C2H6O) to moles.
To convert the mass to moles, we need to divide the given mass (50g) by the molar mass of ethanol (C2H6O).
The molar mass of C2H6O is calculated by multiplying the atomic masses of carbon (C), hydrogen (H), and oxygen (O) by their respective subscripts and adding them together.
Carbon: 12.01 g/mol x 2 = 24.02 g/mol
Hydrogen: 1.01 g/mol x 6 = 6.06 g/mol
Oxygen: 16.00 g/mol x 1 = 16.00 g/mol
Molar mass of C2H6O = 24.02 g/mol + 6.06 g/mol + 16.00 g/mol = 46.08 g/mol
To convert grams to moles: Moles = Mass (g) / Molar Mass (g/mol)
Moles of C2H6O = 50g / 46.08 g/mol = 1.086 mol
Step 2: Determine the stoichiometry (molar ratio) between ethanol and carbon dioxide.
According to the balanced chemical equation, for every 1 mole of ethanol (C2H6O), 3 moles of carbon dioxide (CO2) are produced.
So, 1 mole of C2H6O = 3 moles of CO2
Step 3: Calculate the moles of carbon dioxide produced.
Moles of CO2 = Moles of C2H6O x (3 moles of CO2 / 1 mole of C2H6O)
Moles of CO2 = 1.086 mol x (3 mol CO2 / 1 mol C2H6O) = 3.258 mol CO2
Step 4: Use the ideal gas law to find the volume of carbon dioxide.
The ideal gas law is given by the formula: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Before using the ideal gas law, we need to convert the temperature from Celsius to Kelvin:
Temperature in Kelvin = Temperature in Celsius + 273.15
T = 25°C + 273.15 = 298.15 K
Now we can rearrange the ideal gas law formula to solve for the volume (V):
V = (nRT) / P
Substituting the given values:
V = (3.258 mol x 0.0821 L·atm/mol·K x 298.15 K) / 100 kPa
Note: The ideal gas constant (R) can be expressed in different units. In this case, we are using liters, atmospheres, moles, and Kelvin (L·atm/mol·K).
Step 5: Convert the pressure from kilopascals (kPa) to atmospheres (atm).
1 atmosphere (atm) = 101.325 kilopascals (kPa)
V = (3.258 mol x 0.0821 L·atm/mol·K x 298.15 K) / (100 kPa x 1 atm / 101.325 kPa)
V = 6.391 L
Therefore, approximately 6.391 liters of carbon dioxide (CO2) will be produced.