If 260 mL of O2 gas is collected at 21 degrees Celsius and 101.3 kPa, what volume would this gas occupy at STP?

241.1 mL.

Explanation:

From the general law of ideal gases:

PV = nRT.

where, P is the pressure of the gas.

V is the volume of the container.

n is the no. of moles of the gas.

R is the general gas constant.

T is the temperature of the gas (K).

For the same no. of moles of the gas at two different (P, V, and T):

P₁V₁/T₁ = P₂V₂/T₂.

P₁ = 101.3 kPa = 1.0 atm, V₁ = 260.0 mL, T₁ = 21°C + 273 = 294.0 K. P₂ = 1.0 atm (standard P), V₂ = ? mL, T₂ = 0.0°C + 273 = 273.0 K (standard T).

∴ V₂ = (P₁V₁T₂) / (T₁P₂) = (1.0 atm) (260.0 mL) (273.0 K) / (294.0 K) (1.0 atm) = 241.1 mL.