1. Convert grams to moles:
* The molar mass of oxygen gas (O₂) is 32 g/mol (16 g/mol for each oxygen atom).
* Divide the mass of oxygen gas by its molar mass: 11.3 g / 32 g/mol = 0.353 moles
2. Use the Ideal Gas Law:
* The Ideal Gas Law is PV = nRT, where:
* P = pressure (we'll assume standard pressure, 1 atm)
* V = volume (what we want to find)
* n = number of moles (0.353 moles)
* R = ideal gas constant (0.0821 L·atm/mol·K)
* T = temperature (we'll assume standard temperature, 273 K)
3. Solve for Volume (V):
* Rearrange the Ideal Gas Law: V = nRT/P
* Plug in the values: V = (0.353 mol)(0.0821 L·atm/mol·K)(273 K) / (1 atm)
* Calculate: V ≈ 8.01 L
Therefore, approximately 8.01 liters of oxygen gas are present in 11.3 grams of oxygen gas at standard temperature and pressure.
