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What is the volume of 12.0 grams of oxygen gas at STP

Atomic mass:O = 15.99 grams/moles

1 Answer

7 votes

Answer:

16.82 L.

Step-by-step explanation:

  • We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 1.0 atm, STP conditions).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = mass/molar mass = (12.0 g)/(15.99 g/mol) = 0.7505 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 0.0°C + 273 = 273.0 K, STP conditions).

∴ V = nRT/P = (0.7505 mol)(0.0821 L.atm/mol.K)(273.0 K)/(1.0 atm) = 16.82 L.

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