Question

How many grams of solid KClO₃ are needed to prepare 18.0 L of oxygen, O₂, which is collected over water at 1.00 atm pressure and at a temperature of 22°C? The vapour pressure of water at 22°C is 19.8 torr. The reaction is 2KClO3 (s) --> 2KCl (s) + 3O2 (g)

Answer 1

The balanced reaction is as follows:

2KClO₃(s) 2KCl(s) + 3O₂(g)

Volume of oxygen gas(V) = 18.0 L

Temperature(T) = 22^oC + 273.15 = 295.15 K

Total pressure = 1.00 atm x ( 760 Torr/1 atm) = 760 Torr

The vapor pressure of water at 22^oC = 19.8 Torr

Determine the pressure of oxygen gas as follows:

Total pressure = Pressure of oxygen gas + vapor pressure of water

Rearrange the formula for Pressure of oxygen gas as follows:

The pressure of oxygen gas = Total pressure - vapor pressure of water

The pressure of oxygen gas = 760 Torr -19.8 Torr

The pressure of oxygen gas =740.2 Torr x ( 1 atm /760 Torr) = 0.974 atm

Determine the number of moles of oxygen gas as follows:

The ideal gas equation is as follows:

PV = nRT

Here, the pressure is P, the volume is V, the number of moles is n, Universal gas constant is R, and Temperature in Kelvin is T.

Rearrange the formula for the number of moles as follows:

n = PV/RT

Substitute 0.974 atm for P, 18.0 L for V, 0.08206 L.atm/mol.K for R, and 295.15 K for T and determine the value of number of moles(n) as follows:

n = (0.974 atm x 18.0 L)/(0.08206 L.atm/mol.K x 295.15 K)

n = 0.7239 mol oxygen gas

Now, use the moles of oxygen gas and the balanced chemical reaction and determine the number of moles of KClO₃ as follows:

Mol KClO₃ = 0.7239 mol O₂ x ( 2 mol KClO₃/3 mol O₂) = 0.4825 mol KClO₃

Use the moles of KClO₃ and the molar mass of KClO₃ and determine the mass of KClO₃ as follows:

= 0.4825 mol KClO₃ x (122.55 g KClO₃ / 1 mol KClO₃)

= 59.1 g KClO₃

Thus, 59.1 g KClO₃ solid KClO₃ is needed to prepare the oxygen gas.