Question

A 6.39 g sample of NO and a 7.93 g sample of SO₂ react in a closed 4.74 L container at 399 K, according to the following balanced chemical equation:

2NO(g) + 2SO₂(g) → N₂(g) + 2SO₃(g)

Calculate the P_N2 (in atm) in the container after the reaction has gone to completion.

Answer 1

2NO(g) + 2SO2(g) → N2(g) + 2SO3(g)

Mass of NO = 6.39 g

Molar mass of NO = 30.01 g / mole

Moles of NO = 6.39 / 30.01

= 0.21

Mass of SO2 = 7.93

Molar mass of SO2 = 64.066 g /mole

Moles of SO2 = 7.93 / 64.066

= 0.123

As 2 moles No reacts with 2 moles SO2.

So, SO2 is the limiting reagent.

Moles of N2 produced = moles of So2 / 2

= 0.123 / 2

= 0.061

volume of container = 4.74 L

Temperature = 399 K

Using ideal gas equation:

PV = nRT

P = nRT / V

= 0.061*0.0821*399 / 4.74

P = 0.42 atm.

Hence,

PN2 = 0.42 atm