calculate the volume of oxygen required for complete combustion of 12...
Calculation of Volume of Oxygen Required for Complete Combustion of Ethane Gas at NTP
Formula Used:
The balanced chemical equation for the combustion of ethane gas is:
C2H6 + 3.5O2 → 2CO2 + 3H2O
From the equation, it can be observed that 1 mole of ethane requires 3.5 moles of oxygen for complete combustion.
The volume of oxygen required can be calculated using the ideal gas law:
V = nRT/P
Where,
V = Volume of gas (in dm3)
n = Number of moles of gas
R = Universal gas constant (0.0821 L atm K-1 mol-1)
T = Temperature of gas (in Kelvin)
P = Pressure of gas (in atm)
Solution:
Given, volume of ethane gas = 12.5 dm3
At NTP, temperature (T) = 273 K and pressure (P) = 1 atm
First, let's calculate the number of moles of ethane:
1 mole of any gas occupies 22.4 dm3 at NTP
Therefore, number of moles of ethane = volume of ethane gas / 22.4
Number of moles of ethane = 12.5 / 22.4 = 0.558 moles
Now, let's calculate the number of moles of oxygen required for complete combustion of ethane:
Number of moles of oxygen = 3.5 x number of moles of ethane
Number of moles of oxygen = 3.5 x 0.558 = 1.953 moles
Finally, let's calculate the volume of oxygen required:
V = nRT/P
V = (1.953 x 0.0821 x 273) / 1
V = 44.6 dm3
Therefore, the volume of oxygen required for complete combustion of 12.5 dm3 ethane gas at NTP is 44.6 dm3.