22.4/2 liters.
what is the volume 16 grams of O2 at STP?
To find the volume of 16 grams of O2 at STP (Standard Temperature and Pressure), we can use the Ideal Gas Law Equation: PV = nRT.
Here's how to get the answer:
1. Convert the given mass of O2 from grams to moles.
- To convert grams to moles, divide the given mass by the molar mass of O2, which is 32 g/mol.
- In this case, 16 grams of O2 divided by 32 g/mol equals 0.5 moles of O2.
2. Determine the STP conditions.
- STP is defined as a temperature of 273.15 K (0 degrees Celsius) and a pressure of 1 atmosphere (atm).
3. Substitute the values into the Ideal Gas Law Equation: PV = nRT.
- P represents the pressure, V represents the volume, n represents the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.
- In this case, the pressure (P) is 1 atm, the number of moles (n) is 0.5 moles, the ideal gas constant (R) is 0.0821 L·atm/mol·K, and the temperature (T) is 273.15 K.
- Plug these values into the equation: (1 atm)·(V) = (0.5 moles)·(0.0821 L·atm/mol·K)·(273.15 K).
4. Solve for V (volume).
- Rearrange the equation to solve for V by dividing both sides by 1 atm (pressure): V = (0.5 moles)·(0.0821 L·atm/mol·K)·(273.15 K) / (1 atm).
- Calculate the volume: V = 11.18 liters (rounded to two decimal places).
Therefore, the volume of 16 grams of O2 at STP is approximately 11.18 liters.