AIR CONTAIN 20% oxygen by volume . the volume of air required to oxidise 1gram CNG AT STP WILL BE WHAT
To calculate the volume of air required to oxidize 1 gram of CNG (Compressed Natural Gas) at STP (Standard Temperature and Pressure), we need to know the composition of CNG and the stoichiometric ratio of its combustion reaction.
CNG mainly consists of methane (CH4), which is its primary component. The combustion reaction of methane with oxygen (O2) produces carbon dioxide (CO2), water (H2O), and releases energy.
The balanced chemical equation for the combustion of methane is:
CH4 + 2O2 -> CO2 + 2H2O
From the equation, you can see that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water.
Now let's calculate the stoichiometric quantity of methane and oxygen required. The molar mass of CH4 = 12.01 g/mol (carbon) + 1.01 g/mol (hydrogen) x 4 = 16.05 g/mol. We can consider that 1 gram of CNG contains 1 gram/ 16.05 g/mol = 0.0623 moles of methane.
According to the stoichiometry of the equation, 1 mole of methane reacts with 2 moles of oxygen. So, 0.0623 moles of methane will require 0.0623 moles x 2 = 0.1246 moles of oxygen.
To find the volume of any gas at STP, we can use the molar volume, which is 22.4 liters/mol at STP.
Now, let's calculate the volume of oxygen needed in liters. 0.1246 moles x 22.4 liters/mol = 2.79 liters.
Therefore, the volume of air required to oxidize 1 gram of CNG at STP is approximately 2.79 liters.