Question

Solid potassium chlorate decomposes into solid potassium chloride and oxygen gas. If 2.00 g potassium chlorate decomposes in a chamber with a fixed volume of 0.800 L, and temperature of 25.0◦C. What is the final chamber pressure? Assume all the potassium chlorate decomposes. Express your answer in atmospheres.

Answer 1

Answer : The final chamber pressure is 0.746 atm.

Step-by-step explanation:

First we have to calculate the moles of
`KClO_3`.

Molar mass of
`KClO_3` = 122.5 g/mole

`\text{ Moles of }KClO_3=\frac{\text{ Mass of }KClO_3}{\text{ Molar mass of }KClO_3}=(2.00g)/(122.5g/mole)=0.0163moles`

Now we have to calculate the moles of
`O_2`.

The balanced chemical reaction will be:

`2KClO_3\rightarrow 2KCl+3O_2`

From the balanced reaction we conclude that,

As, 2 moles of
`KClO_3` react to give 3 moles of
`O_2`

So, 0.0163 moles of
`KClO_3` react to give
`(3)/(2)* 0.0163=0.0244` moles of
`O_2`

Now we have to calculate the pressure of gas.

Using ideal gas equation:

`PV=nRT`

where,

P = pressure of gas = ?

V = volume of gas = 0.800 L

T = temperature of gas =
`25.0^oC=273+25.0=298K`

R = gas constant = 0.0821 L.atm/mole.K

n = number of moles of gas = 0.0244 mole

Now put all the given values in the ideal gas equation, we get:

`P* (0.800L)=0.0244mole* (0.0821L.atm/mole.K)* (298K)`

`P=0.746atm`

Therefore, the final chamber pressure is 0.746 atm.