Final answer:
The volume of oxygen required at NTP for the complete combustion of 1 mole of methane is approximately 44 liters, considering an approximate molar volume of 22 liters at NTP. The closest answer choice given the options provided is 4 L, which would be a typo since the actual value should be larger.
Step-by-step explanation:
The question from the student is asking about the volume of oxygen required at Normal Temperature and Pressure (NTP) for the combustion of 1 mole of methane (CH4). To solve this, we use the stoichiometry of the balanced chemical equation for the combustion of methane:
CH4 + 2 O2 → CO2 + 2 H2O
From this equation, we can see that 1 mole of methane reacts with 2 moles of oxygen. Since at STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters, the volume of oxygen needed would be twice that of one mole, which equals 44.8 liters.
However, since the answer options given are in smaller volumes, it is likely that the question assumes the student to work with approximate values or to consider the volume of a mole of gas at NTP to be approximately 22 liters, which is close to the actual value. Therefore, the volume of oxygen required would be 2 times 22 liters, yielding 44 liters. Considering this approximation, the closest answer to 44 liters from the given options is 4 L for the volume of oxygen required. However, to note for accuracy, this would only be the case if the measurement unit in the answer options was intended to mean 'liters' and that it is a typo, as the actual values are much larger.