what volume of air containing 20% O2 will be required to burn completely 10cm3 each of methane and acetylene?
To determine the volume of air needed to burn methane and acetylene completely, we first need to understand their combustion reactions. The balanced chemical equations for the combustion of methane (CH4) and acetylene (C2H2) are as follows:
Methane Combustion:
CH4 + 2O2 -> CO2 + 2H2O
Acetylene Combustion:
2C2H2 + 5O2 -> 4CO2 + 2H2O
From these reactions, we can see that 1 molecule of methane requires 2 molecules of oxygen (O2) to burn completely, while 2 molecules of acetylene need 5 molecules of oxygen.
Now, let's calculate the volume of oxygen required for the combustion of methane and acetylene:
Volume of Oxygen required for methane:
Since 1 molecule of methane requires 2 molecules of oxygen,
The volume of oxygen = (2/1) x volume of methane
Volume of Oxygen required for acetylene:
Since 2 molecules of acetylene require 5 molecules of oxygen,
The volume of oxygen = (5/2) x volume of acetylene
Given that the volume of methane is 10cm3, and the volume of acetylene is also 10cm3, we can substitute these values into the formulas above to determine the required volume of oxygen.
For methane:
Volume of oxygen for methane = (2/1) x 10cm3 = 20cm3
For acetylene:
Volume of oxygen for acetylene = (5/2) x 10cm3 = 25cm3
Therefore, the total volume of oxygen required to completely burn both methane and acetylene is 20cm3 + 25cm3 = 45cm3.
Since the air contains 20% oxygen (O2), we can determine the volume of air needed to provide 45cm3 of oxygen:
Volume of air = (45cm3 / 20%) = 225cm3
Hence, 225cm3 of air containing 20% O2 would be required to burn completely 10cm3 each of methane and acetylene.