What volume will 12.0 g of oxygen gas occupy at 25 c and a pressure of 52.7 kpa?

We can use the ideal gas law equation to find the volume occupied by oxygen gas

PV = nRT

where;

P - pressure - 52.7 kPa

V - volume

n - number of oxygen moles - 12.0 g / 32 g/mol = 0.375 mol

R - universal gas constant - 8.314 Jmol⁻¹K⁻¹

T - temperature - 25 °C + 273 = 298 K

substituting the values in the equation

52 700 Pa x V = 0.375 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K

V = 17.6 L

volume of the gas is 17.6 L