Question

As a result of the reaction, 34.1 grams of ammonia (NH3) was produced. What volume of hydrogen gas at 20.0°C and 1.5 atm was required for this mass of ammonia to be produced, assuming sufficient nitrogen to react?

Answer 1

The volume of H₂ produced is 32.03L

Step-by-step explanation:

The reaction is this:

N₂ + 3H₂ → 2NH₃

Ratio between H₂ and ammonia is NH₃ is 2:3 so the rule of three will be

2 moles from ammonia are produced by 3 moles of hydrogen

My moles of ammonia were produced by ....

I have the mass of the produced ammonia so:

Mass / Molar mass = Moles

34.1 g / 17 g/m = 2 moles

(2 .3) / 2 = 3 moles of Hydrogen were necessary to produce 34.1 g

Apply the Ideal Gases Law equation:

P .V = n . R . T

1.5 atm . V = 2 moles . 0.082 . 293K)

V = (2 moles . 0.082 . 293K) / 1.5atm → 32.03L