Find the volume of 56.0 grams of O2

To find the volume of 56.0 grams of O2, we will use the ideal gas law formula and the molar mass of oxygen.

Firstly, we need to determine the number of moles of O2, using the formula:

moles = mass / molar mass

The molar mass of O2 (oxygen molecule) is 32.0 grams/mole (since the atomic mass of an oxygen atom is approximately 16.0 atomic mass units and O2 contains two oxygen atoms).

So, moles of O2 = 56.0 grams / 32.0 grams/mole = 1.75 moles

Now, we use the ideal gas law to find the volume, which is:

PV = nRT

Where: - P is the pressure - V is the volume - n is the number of moles - R is the ideal gas constant - T is the absolute temperature (in Kelvin)

We need to assume standard temperature and pressure for this calculation unless specified otherwise. Standard temperature and pressure (STP) is 0°C (273.15 K) and 1 atmosphere (atm).

Using the ideal gas constant R = 0.0821 L·atm/mol·K, the volume V at STP can be calculated by rearranging the ideal gas law:

V = nRT / P

If we assume STP conditions (P = 1 atm, T = 273.15 K):

V = (1.75 moles) * (0.0821 L·atm/mol·K) * (273.15 K) / (1 atm) V = (1.75) * (0.0821) * (273.15) V ≈ 39.33 liters

Therefore, the volume of 56.0 grams of O2 at STP is approximately 39.33 liters.