Final answer:
To calculate the volume of sulfur trioxide gas formed at RTP from 9.6 g of oxygen, we first convert the mass of oxygen to moles, and then use the molar volume of gas at RTP to find the volume. However, the calculated volume does not match the provided options, indicating a possible error in the question or calculations.
Step-by-step explanation:
The question involves calculating the volume of sulfur trioxide gas formed at room temperature and pressure (RTP) when 9.6 grams of oxygen react with sulfur dioxide. To determine this, we first need to find the molar ratio of oxygen to sulfur trioxide using the balanced chemical equation for their reaction.
The balanced chemical equation for the reaction between sulfur dioxide (SO2) and oxygen (O2) to form sulfur trioxide (SO3) is:
2 SO2 (g) + O2 (g) → 2 SO3 (g)
This shows us that 1 mole of O2 is required to produce 2 moles of SO3. Next, we convert the mass of oxygen to moles, knowing that the molar mass of O2 is approximately 32.00 g/mol:
Moles of O2 = 9.6 g / 32.00 g/mol = 0.3 moles
Since the stoichiometric coefficients indicate that 1 mole of O2 produces 2 moles of SO3, 0.3 moles of O2 will produce 0.6 moles of SO3. At RTP, one mole of a gas occupies 22.4 liters; therefore, the volume of SO3 produced will be:
Volume of SO3 = 0.6 moles × 22.4 L/mol = 13.44 L
However, this answer does not match any of the provided options (A) 22.4 L (B) 11.2 L (C) 5.6 L (D) 2.8 L. Therefore, we must assume that there was a mistake in the question or in the process of calculation. Always ensure to double-check calculations and the given options in the question.