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30 December, 20:57

What volume does 43.5 g of n2 occupy at stp? (r = 0.08206 l⋅atm/mol⋅k) ?

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Answers (2)
  1. 30 December, 21:34
    0
    Answer is: volume of nitrogen is 34.72 liters.

    m (N₂) = 43.5 g.

    n (N₂) = m (N₂) : M (N₂).

    n (N₂) = 43.5 g : 28 g/mol.

    n (N₂) = 1.55 mol, amount of substance.

    T = 273 K, standard temperature.

    p = 1 atm, standard pressure.

    R = 0.08206 L·atm/mol·K, universal gas constant.

    Ideal gas law: p·V = n·R·T.

    V = n·R·T / p.

    V (N₂) = 1.55 mol · 0.08206 L·atm/mol·K · 273 K / 1.00 atm.

    V (O ₂) = 34.72 L.
  2. 30 December, 23:14
    0
    We are going to use this formula:

    PV = n RT

    when at STP

    p is the pressure = 1 atm

    and n is the moles = mass / molar mass = 43.5 g / 28g/mol

    = 1.56 moles

    R is the ideal gas constant = 0.0821

    and T is a temperature in Kelvin = 273 K

    by substitution, we will get V (the volume)

    1 atm * V = 1.56 moles * 0.0821 * 273K

    ∴ V = 34.96 L
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