how much volume of oxygen at stp in litres is required to burn 4g of methane gas completely
CH4 + 2O2 ==> CO2 + 2H2O
mols CH4 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols CH4 to mols O2.
Now convert mols O2 to L O2 by mols O2 x 22.4 L/mol = L O2.
64.1
To calculate the volume of oxygen required to burn a given amount of methane gas completely, we need to use the balanced chemical equation for the combustion of methane.
The balanced chemical equation for the combustion of methane is as follows:
CH₄ + 2O₂ → CO₂ + 2H₂O
From the balanced equation, we can see that one molecule of methane requires two molecules of oxygen for complete combustion.
Step 1: Convert the mass of methane to moles.
Using the molar mass of methane (CH₄ = 16.04 g/mol), we can calculate the number of moles of methane gas.
Number of moles of methane = mass of methane (in grams) / molar mass of methane.
Number of moles of methane = 4 g / 16.04 g/mol = 0.249 mol (rounded to three decimal places).
Step 2: Determine the molar ratio of methane to oxygen.
From the balanced chemical equation, we know that the molar ratio of methane to oxygen is 1:2.
Step 3: Calculate the number of moles of oxygen required.
Number of moles of oxygen = (Number of moles of methane) x (Molar ratio of oxygen to methane).
Number of moles of oxygen = 0.249 mol × 2 = 0.498 mol (rounded to three decimal places).
Step 4: Convert moles of oxygen to volume at STP.
At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters.
Volume of oxygen required = (Number of moles of oxygen) × (22.4 L/mol).
Volume of oxygen required = 0.498 mol × 22.4 L/mol = 11.15 L (rounded to two decimal places).
Therefore, approximately 11.15 liters of oxygen at STP are required to burn 4 grams of methane gas completely.