Final answer:
The percentage of oxygen in the air can be calculated using the mass of oxygen that has reacted with methane during combustion. After determining the mass of oxygen present in the products, it is deduced that the percentage of oxygen in the air sample is 20%.
Step-by-step explanation:
To calculate the percentage of oxygen in the air used during the combustion of methane, we use the concept of mass conservation and stoichiometry. The combustion of methane can be described by the chemical equation: CH4 + 2O2 → CO2 + 2H2O. Given that 1.6g of methane (CH4) was burnt and 4.4g of carbon dioxide (CO2) and 3.6g of water (H2O) were produced, we can calculate the mass of oxygen that has reacted:
- Mass of carbon in CO2 = 4.4g x (12/44) = 1.2g
- Mass of oxygen in CO2 = 4.4g - 1.2g = 3.2g
- Mass of hydrogen in H2O = 3.6g x (2/18) = 0.4g
- Mass of oxygen in H2O = 3.6g - 0.4g = 3.2g
- Total mass of oxygen in products = 3.2g + 3.2g = 6.4g
The mass of the air that reacted is 32g, and since it consisted of only methane and oxygen (ignoring other trace gases for this calculation), the mass of oxygen in the air is 6.4g. Therefore, the percentage of oxygen in the air sample is given by:
(Mass of oxygen / Mass of air) x 100% = (6.4g / 32g) x 100% = 20%.
Thus, the correct answer is B) 20%.