If there are 0.68 kg of sulfur trioxide at STP, how many liters does this occupy?

Question

I need help with this question

Answer 1

Use the fact the 1 mole of an ideal gas = 22.4 liters.

Sulfur trioxide = SO3

molar mass SO3 = 64 g/mol

moles SO3 present = 0.68 kg SO3 x 1000 g/kg x 1 mol/64 g = 10.625 moles SOe

Volume occupied = 10.625 moles x 22.4 L / mol = 238 L = 240 L (2 sig. figs.)