<u>Answer:</u> The volume of oxygen gas needed is 156.8 L
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
Given mass of methane = 56 g
Molar mass of methane = 16 g/mol
Putting values in above equation, we get:
The chemical equation for the combustion of methane follows:
By Stoichiometry of the reaction:
1 mole of methane reacts with 2 moles of oxygen gas
So, 3.5 moles of methane will react with = of oxygen gas
<u>At STP:</u>
1 mole of a gas occupies 22.4 L of volume
So, 7 moles of oxygen gas occupies of volume
Hence, the volume of oxygen gas needed is 156.8 L