What volume does 16.0g of o2 occupy at stp
Hello, at STP a standard gas has 22.4 L volume and 1 atm pressure. The atomic mass of one mole of O2 gas would be 32 g. And this is 16 g of 02 which would mean that there is 16g/32g = 0.5 mol O2 gas. So, if 1 mol of any standard gas at STP is 22.4 L then 0.5 mol would be 11.2 Liters. Hope this helps.
11.2 Liters
To determine the volume that 16.0g of O2 occupies at STP (Standard Temperature and Pressure), you need to use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in atmospheres)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (in Kelvin)
At STP, the temperature is 273.15 Kelvin, and the pressure is 1 atmosphere. By rearranging the ideal gas law equation, we can solve for the volume:
V = nRT/P
To determine the number of moles (n), we need to convert the mass of O2 (16.0g) into moles using the molar mass of O2. The molar mass of O2 is 32.00 g/mol (16.00 g/mol per oxygen atom × 2 oxygen atoms).
n = mass/molar mass
n = 16.0g / 32.00 g/mol
n = 0.50 mol
Now, we can substitute the values into the equation and solve for V:
V = (0.50 mol)(0.0821 L·atm/(mol·K))(273.15 K) / 1 atm
V ≈ 11.2 L
Therefore, 16.0g of O2 would occupy approximately 11.2 liters of volume at STP.