What is the volume of oxygen required to bum completely 45cm3 of methane at s.t.p?
Answer Details
When methane burns completely in oxygen, it produces carbon dioxide and water vapor. According to the balanced chemical equation for the combustion of methane:
CH4 + 2O2 → CO2 + 2H2O
From the equation, we see that 1 mole of methane (16g) reacts with 2 moles of oxygen (32g) to give 1 mole of carbon dioxide (44g) and 2 moles of water vapor (36g).
At s.t.p, 1 mole of any gas occupies 22.4 liters or 22,400 cm3. Therefore, the volume of 45cm3 of methane is equal to 45/22,400 = 0.002 moles.
Since methane reacts with oxygen in a 1:2 mole ratio, we need twice the number of moles of oxygen to react completely with the given amount of methane. Thus, we need 2 x 0.002 = 0.004 moles of oxygen.
At s.t.p, 1 mole of oxygen occupies 22.4 liters or 22,400 cm3. Therefore, the volume of 0.004 moles of oxygen required is 0.004 x 22,400 = 89.6 cm3.
Therefore, the correct option is 90.0cm3, which is closest to the calculated volume of oxygen required.