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60.0 kL of hydrogen at SATP (24.8 L/mol) is reacted with excess amount of nitrogen. The resulting
ammonia gas needs to be stored in a gas tank with a volume of 6.25 kL and under 2000 kPa of pressure.
What should the temperature regulator of the gas tank be set at?

Respuesta :

mickymike92

The temperature regulator of the gas tank be set at 842.64 K

What volume of ammonia gas is produced by reacting 60.kL of hydrogen with excess nitrogen at SATP?

The volume of hydrogen ammonia produced is given by the mole ratio of the gases in the equation below:

3 H₂ + N₂ → 2 NH₃

The mole ratio is 2:3

Volume of ammonia produced = 2/3 * 60 kL = 40 kL

The temperature regulator value is determined using the general gas equation:

  • P₁V₁/T₁ = P₂V₂/T₂

T₂ = P₂V₂T₁/P₁V₁

  • P₂ = 2000 kPa
  • V₂ = 6.25 kL
  • T₁ = 273.15 K
  • P₁ = 101.3 kPa
  • V₁ = 40 kL

T₂ = (2000 * 6.25 * 273.15)/(101.3 * 40)

T₂ = 842.64 K

The temperature regulator of the gas tank be set at 842.64 K

In conclusion, the general gas equation is useful in determining gas volume, pressure and temperatures.

Learn more about general gas equation at: https://brainly.com/question/4147359

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