What volume of oxygen gas is needed for the complete combustion of 4.00 L of ethane gas, C2H6? Assume that the temperature and pressure of the reactants are the same.
2C2H6 + 7O2 ==> 4CO2 + 6H2O
When gases are involved one can take a shortcut and use liters and mols interchangeably.
So 4.00 L C2H6 x (7 mols O2/2 mols C2H6) = 4.00 x 7/2 = ? L O2
Is not there in my options
To determine the volume of oxygen gas needed for the complete combustion of ethane gas (C2H6), we need to balance the chemical equation first:
C2H6 + O2 → CO2 + H2O
The balanced equation shows that 1 mole of ethane reacts with 3.5 moles of oxygen gas to produce 2 moles of carbon dioxide and 3 moles of water.
Since the volume of a gas is directly proportional to the number of moles, we can use the balanced equation to calculate the volume ratio between ethane and oxygen.
Let's calculate:
1. Find the number of moles of ethane gas (C2H6) present in 4.00 L:
Using the ideal gas law equation PV = nRT, we can calculate the moles of ethane gas.
Assuming temperature (T) and pressure (P) are constant, re-arranging the equation will give us:
n = PV/RT
Assuming standard temperature and pressure (STP), T = 273 K and P = 1 atm, and R = 0.0821 L·atm/mol·K, we can calculate the number of moles:
n = (4.00 L)(1 atm)/(0.0821 L·atm/mol·K)(273 K)
n ≈ 0.188 mol
2. Determine the stoichiometric ratio between ethane and oxygen:
According to the balanced equation, the ratio is 1:3.5 (1 mole of ethane reacts with 3.5 moles of oxygen).
3. Calculate the number of moles of oxygen gas required:
Number of moles of oxygen = 0.188 mol × 3.5 mol/mol
Number of moles of oxygen ≈ 0.658 mol
4. Convert the number of moles of oxygen gas to volume:
Using the ideal gas law equation, we can calculate the volume of oxygen gas.
V = nRT/P
V = (0.658 mol)(0.0821 L·atm/mol·K)(273 K)/(1 atm)
V ≈ 15.8 L
Therefore, approximately 15.8 L of oxygen gas is needed for the complete combustion of 4.00 L of ethane gas, assuming the temperature and pressure of the reactants are the same.
To calculate the volume of oxygen gas needed for the complete combustion of ethane gas, we need to balance the chemical equation first. The balanced equation for the combustion of ethane is:
C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O
According to the balanced equation, it shows that 1 mole of ethane (C2H6) reacts with 7/2 moles of oxygen gas (O2) to produce 2 moles of carbon dioxide (CO2) and 3 moles of water (H2O).
From the balanced equation, we can see that the stoichiometric ratio of C2H6 to O2 is 1:7/2. This means that for every 1 mole of ethane, we need 7/2 moles of oxygen gas.
To calculate the volume of oxygen gas, we first need to convert the given volume of ethane gas (4.00 L) into moles using the ideal gas law equation:
PV = nRT
We assume the temperature and pressure are the same, so the equation becomes:
V = nRT/P
Where:
V = volume of gas (ethane)
n = number of moles of gas (ethane)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
P = pressure (in atm)
Once we find the number of moles of ethane, we can use the stoichiometric ratio to find the number of moles of oxygen needed. Finally, we will use the ideal gas law equation again to convert the number of moles of oxygen into volume, similar to what we did for ethane.
Here are the step-by-step calculations:
Step 1: Convert the given volume of ethane gas to moles.
We'll assume the temperature and pressure are the same as stated, but these values are not provided in the question. For this explanation, let's assume the temperature is 298 K and the pressure is 1 atm.
V(C2H6) = 4.00 L
T = 298 K
P = 1 atm
R = 0.0821 L·atm/(mol·K)
Using the ideal gas law equation:
n(C2H6) = PV/RT
n(C2H6) = (1 atm * 4.00 L) / (0.0821 L·atm/(mol·K) * 298 K)
n(C2H6) ≈ 0.192 moles (rounded to three decimal places)
Step 2: Convert moles of ethane to moles of oxygen.
According to the balanced equation, for every 1 mole of ethane, 7/2 moles of oxygen are needed.
n(O2) = (7/2) * n(C2H6)
n(O2) = (7/2) * 0.192 moles
n(O2) ≈ 0.672 moles (rounded to three decimal places)
Step 3: Convert moles of oxygen to volume.
Assuming the temperature and pressure are the same as before:
n(O2) = 0.672 moles
R = 0.0821 L·atm/(mol·K)
T = 298 K
P = 1 atm
Using the ideal gas law equation:
V(O2) = nRT/P
V(O2) = (0.672 mol * 0.0821 L·atm/(mol·K) * 298 K) / 1 atm
V(O2) ≈ 13.27 L (rounded to two decimal places)
Therefore, the volume of oxygen gas needed for the complete combustion of 4.00 L of ethane gas is approximately 13.27 L.