how many liters of oxygen are necessary for the combustion of 425 g methane assuming the reaction takes place at STP
To determine the number of liters of oxygen required for the combustion of methane at STP (Standard Temperature and Pressure), we need to follow a few steps:
Step 1: Write the balanced chemical equation for the combustion of methane.
The balanced chemical equation for the combustion of methane (CH₄) is as follows:
CH₄ + 2O₂ -> CO₂ + 2H₂O
Step 2: Determine the molar mass of methane and oxygen.
The molar mass of methane (CH₄) can be calculated as follows:
(Mass of Carbon + 4 x Mass of Hydrogen) = (12.01 g/mol + 4 x 1.01 g/mol) = 16.04 g/mol
The molar mass of oxygen (O₂) is simply 32.00 g/mol (16.00 g/mol x 2).
Step 3: Convert the given mass of methane into moles.
To convert the mass of methane into moles, divide the given mass by the molar mass:
425 g / 16.04 g/mol ≈ 26.5 mol
Step 4: Use the balanced equation to determine the molar ratio between methane and oxygen.
Based on the balanced equation, we know that for every 1 mole of methane, 2 moles of oxygen are required.
Step 5: Calculate the number of moles of oxygen needed.
To calculate the moles of oxygen required, multiply the number of moles of methane by the mole ratio:
26.5 mol CH₄ x (2 mol O₂ / 1 mol CH₄) = 53 mol O₂
Step 6: Convert the number of moles of oxygen into liters.
At STP, 1 mole of gas occupies 22.4 liters. Therefore, we can calculate the volume of oxygen using the following conversion factor:
53 mol O₂ x (22.4 L / 1 mol) = 1,187.2 L
So, approximately 1,187.2 liters of oxygen are necessary for the combustion of 425 g of methane at STP.