What mass, in grams, of oxygen gas (o2) is contained in a 8.5-liter tank at 32.0 degrees celsius and 3.23 atmospheres?

For this problem we assume that oxygen is an ideal gas. So, we use the equation PV=nRT where P is pressure, V is the volume, n is the number of moles, R is a universal constant and T is the temperature. We first solve for the number of moles n. Then, using the molar mass of oxygen we convert it to grams.

PV=nRT

n = PV / RT

n = 3.23 (8.5) / 0.08206 (32+273.15)

n = 1.0964 mol

mass = 1.0964 mol (32g / 1 mol) = 35.09 g O2