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 data:

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

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