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:

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).

P₁V₁/T₁ = P₂V₂/T₂.

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