Question

12 liters of ammonia (NH3) weighs how much at room temperature and pressure (293 K and 100 kPa)?

• O A. 12 grams
• O B. 0.5 grams
• O C. 17 grams
• O D. 8.5 grams

Answer 1

• Option D) 8.5 grams

Step-by-step explanation:

1) Data:

a) V = 12 liter

b) T = 293 K

c) p = 100 kPa

d) mass = ?

2) Formulae:

a) Ideal gas equation: pV = nRT

b) molar mass = mass in grams / number of moles

3) Solution:

a) Calculate the number of moles of ammonia, n:

• pV = nRT ⇒ n = RT / (pV)

• R = universal constant of gases = 8.3145 liter-kPa / mol-K

• n = 100 kPa × 12 liter / (293 K × 8.3145 liter-kPa/mol-K) = 0.493 mol

b) Convert moles to grams:

• Molar mass of NH₃ = 14.007 g/mol + 3×1.008 g/mol = 17.03 g/mol

• mass = 0.493 mol × 17.03 g/mol = 8.4 g

Thus, the answer is the option D. 8.5 grams.