Question

If 2.71 moles of nitrogen reacts with 4.26 moles of hydrogen, what is the maximum volume of ammonia (in L) that can be produced at a pressure of 1.18 atm and 306 K. N2(g) + 3 H2(g) → 2 NH3(g)

Answer 1

Reaction take place as follow

3H₂ + N₂  2NH₃

According to reaction 3 mole of H₂ react with 1 mole of nitrogen therefore to react with 2.71 mole of N₂ required mole of H₂ = 2.71 X 3 = 8.13 mole of H₂ But H₂ given only 4.26 therefore H₂ is limitig reactant
According to reaction 3 mole of H₂ produce 2 mole of NH₃ then 4.26 H₂ produce NH₃ = 4.26 X 2 / 3 = 2.84 mole

2.84 mole of NH₃ produce

use ideal gas equation to calculate volume of NH₃ gas

We know that PV = nRT

V = nRT/P

n = 2.84 mole,

T = 306 K,

P= 1.18 atm,

R = 0.08205 L atm mol^-1 K^-1 ( R = gas constant)

V = ?

Substitute these value in above equation.

V = 2.84 X0.08205 X306 / 1.18 = 71.31 L

71.31 liter of NH₃ gas produced