Question

calculate the volume of O2 gas at STP required to burn 30g of CH4gas as shown in the following reaction​

Answer 1

Answer:

Consider the combustion reaction of CH4 with O2,

CH4 + 2 O2 → CO2 + 2H2O

Here, 1 mole of CH4 needs 2 moles of molecular O2.

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.