Answer:
From the balanced chemical equation, we know that 1 mole of methane reacts with 2 moles of oxygen gas to produce 1 mole of carbon dioxide and 2 moles of water vapor. At STP (standard temperature and pressure, 0°C and 1 atm), 1 mole of gas occupies 22.4 L of volume.
First, we need to determine how many moles of methane and oxygen are involved in the reaction. Since the volumes of gases are given at STP, we can use the ideal gas law:
n = PV/RT
where n is the number of moles of gas, P is the pressure, V is the volume, R is the gas constant, and T is the temperature. At STP, the pressure is 1 atm and the temperature is 0°C or 273 K. Therefore, we have:
n(CH4) = (29.2 L)(1 atm)/(0.0821 L·atm/mol·K)(273 K) ≈ 1.15 mol
n(O2) = (63.3 L)(1 atm)/(0.0821 L·atm/mol·K)(273 K) ≈ 2.49 mol
From the balanced chemical equation, we know that 1 mole of methane produces 1 mole of carbon dioxide. Therefore, we can determine the volume of carbon dioxide produced using the ideal gas law:
V(CO2) = n(CO2)RT/P
where V is the volume, n is the number of moles, R is the gas constant, T is the temperature, and P is the pressure. At STP, the pressure is 1 atm and the temperature is 0°C or 273 K. Therefore, we have:
V(CO2) = (1.15 mol)(0.0821 L·atm/mol·K)(273 K)/(1 atm) ≈ 26.7 L
Therefore, approximately 26.7 L of carbon dioxide gas is produced in the reaction.