Question

What is the volume occupied by 11.0 g of propane gas, C₃H₈, at STP?

a. 21.6 L
b. 89.6 L
c. 0.250 L
d. 4.00 L
e. 5.59 L

Answer 1

The volume occupied by 11.0 g of propane gas at STP is calculated to be about 5.58 liters, so the closest answer choice is option e) 5.59 L.

Step-by-step explanation:

The question asks for the volume occupied by 11.0 g of propane gas, C3H8, at STP. To answer, we use the molar volume at STP, which is 22.4 liters per mole for any gas. First, we need to calculate the number of moles of propane:

• Propane's molar mass = 3(12.01 g/mol C) + 8(1.008 g/mol H) = 44.096 g/mol.
• Moles of propane = 11.0 g / 44.096 g/mol ≈ 0.249 moles.
• Volume occupied by propane = Moles of propane × Molar volume at STP = 0.249 moles × 22.4 L/mol ≈ 5.58 L.The volume occupied by 11.0 g of propane gas, C₃H₈, at STP can be calculated using the concept of molar volume of gas at STP. The molar volume of a gas at STP is 22.4 L/mol. In this case, we need to determine the number of moles of propane using its molar mass, and then multiply it by the molar volume:
• Molar mass of C₃H₈ = 44.1 g/mol
• Number of moles = mass / molar mass = 11.0 g / 44.1 g/mol = 0.249 mol
• Volume occupied = number of moles * molar volume = 0.249 mol * 22.4 L/mol ≈ 5.59 L
• Therefore, the correct answer is option e. 5.59 L.