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

Question

Cameron Dorsey

Chemistry

9 April, 18:13

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

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Answers (1)

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Maggie Anthony · Answer 1

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.