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How many grams are in 98 L of o2 at stp

1 Answer

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Answer:

139.9 g ≅ 140.0 g.

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, standard P).

V is the volume of the gas in L (V = 98.0 L).

n is the no. of moles of the gas in mol (n = ??? mol).

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

T is the temperature of the gas in K (T = 273 K, standard T).

∴ n = PV/RT = (1.0 atm)((98.0 L)/(0.0821 L.atm/mol.K)(273 K) = 4.37 mol.

  • To find the grams of O₂, we can use the relation:

n = mass/molar mass.

mass of O₂ = (n)*(molar mass) = (4.37 mol)(32.0 g/mol) = 139.9 g ≅ 140.0 g.

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