What is the volume of oxygen required to bum completely 45cm\(^{3}\) of methane at s.t.p?

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.