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.

Explanation:

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

Molar mass of $KClO_3$ = 122.5 g/mole

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

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 $\frac{3}{2}\times 0.0163=0.0244$ 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:

$P\times (0.800L)=0.0244mole\times (0.0821L.atm/mole.K)\times (298K)$

$P=0.746atm$

Therefore, the final chamber pressure is 0.746 atm.