Question

2.0L of hydrogen gas is mixed with 3.0L of nitrogen gas at STP in a rigid 5.0L vessel. A reaction occurs, producing ammonia gas NH3 and causing the pressure to change. How many moles of ammonia are produced? Hint: Write reaction and determine LR first How many moles of excess reactant remain in the vessel after the reaction? What is the final total pressure after the reaction if the temperature is kept at 273 K?

Answer 1

Moles NH₃: 0.0593

0.104 moles of N₂ remain

Final pressure: 0.163atm

Step-by-step explanation:

The reaction of nitrogen with hydrogen to produce ammonia is:

N₂ + 3 H₂ → 2 NH₃

Using PV = nRT, moles of N₂ and H₂ are:

N₂: 1atmₓ3.0L / 0.082atmL/molKₓ273K = 0.134 moles of N₂

H₂: 1atmₓ2.0L / 0.082atmL/molKₓ273K = 0.089 moles of H₂

The complete reaction of N₂ requires:

0.134 moles of N₂ × (3 moles H₂ / 1 mole N₂) = 0.402 moles H₂

That means limiting reactant is H₂. And moles of NH₃ produced are:

0.089 moles of H₂ × (2 moles NH₃ / 3 mole H₂) = 0.0593 moles NH₃

Moles of N₂ remain are:

0.134 moles of N₂ - (0.089 moles of H₂ × (1 moles N₂ / 3 mole H₂)) = 0.104 moles of N₂

And final pressure is:

P = nRT / V

P = (0.104mol + 0.0593mol)×0.082atmL/molK×273K / 5.0L

P = 0.163atm