Air/fuel ratio by weight for combustion of methane with theoretical quantity of air will be about

ANS:B - 17 : 1

The air/fuel ratio by weight for the combustion of methane with the theoretical quantity of air can be calculated based on the stoichiometry of the reaction. The balanced chemical equation for the combustion of methane (CH4​) with oxygen (O2​) is: CH4​+2O2​→CO2​+2H2​O From the balanced equation, we 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. The molar mass of methane (CH4​) is 16g/mol, and the molar mass of oxygen (O2​) is 32g/mol. To find the air/fuel ratio by weight, we can calculate the mass of oxygen required to burn one mole of methane and then express it as a ratio to the mass of methane.

1. Mass of oxygen required to burn one mole of methane: 1 mole of CH4​×2moles of O2/1mole of CH4​​​×​32g of O2​​/ 1mole of O2=64g of O2​
2.  Air/fuel ratio by weight: Mass of O2/Mass of CH4​ ​​=16g/64g​=4

Therefore, the air/fuel ratio by weight for the combustion of methane with the theoretical quantity of air is approximately 4:14:1. However, since air contains only about 21%21% of oxygen, we need to divide the calculated ratio by this percentage to determine the actual air/fuel ratio. Actual air/fuel ratio=40.21≈19Actual air/fuel ratio=0.214​≈19 So, the closest option to the actual air/fuel ratio is approximately 17:1. Therefore, the correct answer is 17:1.