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How many liters will 10.0 grams of hydrogen occupy at stp​

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Answer:

38.36 L.

Step-by-step explanation:

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

where, P is the pressure of H₂ in atm (P = 1.0 atm, STP conditions).

V is the volume of H₂ in L (V = ??? L).

n is the no. of moles of H₂ in mol (n = mass/molar mass = (10.0 g)/(2.0 g/mol) = 5.0 mol).

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

T is the temperature of H₂ in K (T = 0.0°C + 273 = 273.0 K, STP conditions).

∴ V = nRT/P = (5.0 mol)(0.0821 L.atm/mol.K)(273.0 K)/(1.0 atm) = 38.36 L.

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