What is the volume of 128 g of Oxygen at STP?... here's my work...
O=16
128/16= 8 moles
8(22.4) = 179.2 L
Where did i go wrong?
Oxygen is a diatomic molecule. It exists as O2 and not O.
128g/32 = 4.00 moles.
4.00 x 22.4L/mol = 89.6 L.
1 mol of an ideal gas occupies 22.4 L at STP. You have how many moles O2? That is moles = grams/molar mass.
So are u implying that the of moles of O2 is wrong? How do u calculate molar mass?
Many of the gases are diatomic.
H2, N2, O2, F2, Cl2, Br2, I2 (Br and I are not gases though at STP). The noble gases of He, Ne, Ar, Kr, Xe are monatomic. S and P are S8 and P4; they are solids.
Your calculation of the number of moles of oxygen is correct: 128 g / 16 g/mol = 8 moles. However, the mistake occurs in the next step when you multiply the number of moles by 22.4 L/mol.
The value of 22.4 L/mol is the molar volume of any gas at STP (standard temperature and pressure), but it only holds true for ideal gases. Oxygen, however, is not an ideal gas. At STP, the behavior of oxygen deviates from ideality to some extent.
To accurately calculate the volume of 8 moles of oxygen at STP, you need to use the molar volume that specifically applies to oxygen, rather than the general value of 22.4 L/mol.
The molar volume of oxygen at STP is approximately 22.71 L/mol. So, the correct calculation should be:
8 moles × 22.71 L/mol = 181.68 L
Thus, the volume of 128 g of oxygen at STP is approximately 181.68 L.