Question

Ammonia gas is synthesized by combining hydrogen and nitrogen: 3H₂(g) + ϩN₂ (g) -> 2NH₃ (g)

To produce 562g of NH₃, what volume of air (the source of N₂) is required if the air is introduced at 29 C and 99.3kPa? (Assume the air sample has 78.1 mol% N₂)

Answer 1

No.of moles of Ammonia = 562/17 = 33.06,

Pressure = 99.3×1000/101325 = 0.98 atm

(1 atm = 101325 Pa)

Volume of N​​​​​​H₃ = nRT/P = 33.06× 0.0821 × 302/0.98

= 836.4 L

From the chemical reaction,

2 moles Ammonia produced by 1mol Nitrogen

2L Ammonia produced by 1L Nitrogen (According to Avogadro's Law)

836.4 L Ammonia produced by 418.2 L Nitrogen .

Since 100L air contains 78.1L Nitrogen

?    <------------- 418.2 L Nitrogen

= 100×418.2/78.1 = 535.467L