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9 April, 18:13

How many moles of 18L of NH3 at 30 celcius and 912 mmHg? how many grams?

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  1. 9 April, 18:35
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    Answer is: mass of ammonia is 14.76 grams and amount is 0.87 moles.

    V (NH ₃) = 18 L.

    T = 30°C = 303.15 K.

    p = 912 mmHg : 760 mmHg/atm = 1.2 atm.

    R = 0.08206 L·atm/mol·K.

    Ideal gas law: p·V = n·R·T.

    n = p·V / R·T.

    n (NH₃) = 1.2 atm ·18 L / 0.08206 L·atm/mol·K · 303.15 K.

    n (NH₃) = 0.87 mol.

    m (NH₃) = n (NH₃) · M (NH₃).

    m (NH₃) = 0.87 mol · 17 g/mol.

    m (NH₃) = 14.76 g.
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