Question

Upon heating, potassium chlorate decomposes to potassium chloride and oxygen gas. The oxygen gas is collected over water at 14.0 °C in a 500.-mL container. If the total pressure is 766 torr,

How many mol of oxygen gas is in the container?
What is the mole fraction of oxygen gas in the container?.
(Mole fraction = mole of substance of interest / mole of the system)

How many grams of potassium chlorate is decomposed?

Answer 1

Answer –

We are given, potassium chlorate decomposes to potassium chloride and oxygen gas

So reaction –

2KClO₃ -------> 2KCl + 3O₂

For O₂ gas- temperature, T = 14+273 = 287 K , volume = 500 mL = 0.500 L

Total pressure = 766 torr

So we need to calculate first pressure of O₂

We know , total pressure = sum of partial pressures

Total pressure = pressure of O₂ + pressure of water vapor at 14^oC

Pressure of water vapor at 14^oC =12 torr

So, pressure of O₂ = 766 – 12 = 754 torr

Pressure in atm = 754 /760 = 0.992 atm

Now we need to calculate moles of O₂

We need to use ideal gas law

PV = nRT

n = PV/RT

n = 0.992 atm * 0.500 L / 0.0821 L.atm.mol^-1.K^-1 * 287

   = 0.0211 moles of O₂

0.0211 moles of oxygen gas is in the container

Now we need to calculate moles of other product and reactant

From the balanced equation –

3 moles of O₂ = 2 moles of KCl

So, 0.0211 moles of O₂ = ?

= 0.0211 moles of O₂ * 2 moles of KCl / 3 moles of O₂

= 0.0140 moles of KCl

Moles of KClO₃

From the balanced equation –

3 moles of O₂ = 2 moles of KClO₃

So, 0.0211 moles of O₂ = ?

= 0.0211 moles of O₂ * 2 moles of KClO₃ / 3 moles of O₂

= 0.0140 moles of KClO₃

So total moles = 0.0211 + 0.0140 + 0.0140

                       = 0.0491moles

Mole fraction of O₂ = 0.0211 / 0.0491 = 0.429

Now we need to calculate mass of potassium chlorate

We calculated moles of potassium chlorate and we know molar mass of potassium chlorate

So,

Mass of KClO₃ = 0.0140 moles * 122.55 g/mol

                         = 1.72 g