Question

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)

Answer 1

Volume of ammonia 3 L.

Step-by-step 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