Question

the volume of o2 at STP required for the complete combustion of 4 g CH4

Answer 1

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.

Answer 2

Answer:

Answer:You have 4 g of CH4 or 4 g/(16 g/mole) = 0.25 mole of CH4. Since mole ratio between CH4 and O2 = 1 : 2, then, 0.25 mole of CH4 stoichiometrically reacts with 2/1 x 0.25 mole of O2 = 0.50 mole of O2. Thus, the volume of O2 required to completely combust CH4 at STP condition = 0.50 mole x 22.4 L/mole = 11.20 L.