Grade 11Physical Chemistry
How much volume of oxygen at STP in litres is required to burn 4 g of methane gas completely?
2 Answers
Arun
Approved Tutor Answer8 Years agoDear student
The balanced chemical equation for combustion of methane is given as:
CH4 + 2O2 → CO2 + 2H2O
So 1 mole of CH4 requires 2 moles of O2 for complete combustion.
Molar mass of methane = 16 g.
Number of moles of methane in 4 g sample = 4 g/16 g = 1/4 mole.
If 1 mole of CH4 requires 2 moles of O2 for complete combustion.
Therefore 1/4 mole of CH4 will requires = 1/4 ×2 moles of O2 = 0.5 mol of O2 for complete combustion.
1 mole of all the gases occupy 22.4 L at STP
Therefore 0.5 mole will occpy 22.4 L × 0.5 mol = 11.2 L
So 11.2 L of O2 is required for complete combustion of 4 g CH4.
CH4 + 2O2 → CO2 + 2H2O
So 1 mole of CH4 requires 2 moles of O2 for complete combustion.
Molar mass of methane = 16 g.
Number of moles of methane in 4 g sample = 4 g/16 g = 1/4 mole.
If 1 mole of CH4 requires 2 moles of O2 for complete combustion.
Therefore 1/4 mole of CH4 will requires = 1/4 ×2 moles of O2 = 0.5 mol of O2 for complete combustion.
1 mole of all the gases occupy 22.4 L at STP
Therefore 0.5 mole will occpy 22.4 L × 0.5 mol = 11.2 L
So 11.2 L of O2 is required for complete combustion of 4 g CH4.
Regards
Arun (askIITians forum expert)
Janki upadhyay
CH4 + O2 = Co2 + H2O
Balanced equation
CH4 +2O2 = CO2 + 2H2O
2 moles of O2 is required for complete combustion of 1 mole CH4
Moles of CH4 = given mass / molar mass
4g/16g=1/4 moles
Moles of O2 Required for complete combustion of 1/4 moles =1/4×2=0.5 moles
At stp one mole of all gases = 22.4 litres
Thus 0.5 moles of O2 = 11.2 litres
Thus volume of O2 required for complete combustion of 4 g CH4 is 11.2 litres
Hope it helped you
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