Under laboratory conditions of 25.0 degrees C and 99.5 kPa, what is the maximum number of liters of ammonia that could be produced from 1.50 L of nitrogen according to the following equation?

N2 (g) + 3H2 (g) - --> 2NH3 (g)

Volume of ammonia 3 L.

Explanation:

Given dа ta:

Temperature = 25°C (25+273 = 298 k)

Pressure = 99.5 kpa (99.5/101 = 0.98 atm)

Volume of nitrogen = 1.50 L

Volume of ammonia = ?

Solution:

Chemical equation:

N₂ + 3H₂ → 2NH₃

Moles of nitrogen:

PV = nRT

n = PV/RT

n = 0.98 atm * 1.50 L / 0.0821 atm. L/mol. K * 298 K

n = 1.47 / 24.5 / mol

n = 0.06 mol

Now we will compare the moles of nitrogen with ammonia.

N₂ : NH₃

1 : 2

0.06 : 2*0.06 = 0.12 mol

Volume of ammonia:

PV = nRT

V = nRT/P

V = 0.12 mol * 0.0821 atm. L/mol. K * 298 K / 0.98 atm

V = 2.9 atm. L / 0.98 atm

V = 3 L