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

= 0.213

Mass of SO2 = 7.93

Molar mass of SO2 = 64.06 g /mole

Moles of SO2 = 7.93 / 64.06

= 0.123

As 2 moles NO reacts with 2 moles SO2.

So, 0.213 mole NO will reacts with 0.213 moles of SO2.

but we have only 0.123 moles of 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