Question

2. Determine the volume in liters occupied by 120 g of O2 gas at STP.

Answer 1

120 g of O2 is equal to approximately 120/32 = 3.75 mol O2. At STP, one mole of any gas occupies 22.4 L. Thus, at STP, 3.75 mol of O2 gas would occupy 3.75 • 22.4 = 84 L.