Answer :
Answer : The volume of [tex]O_2[/tex] gas produced is 2.24 L
Explanation :
First we have to calculate the moles of [tex]KClO_3[/tex].
[tex]\text{Moles of }KClO_3=\frac{\text{Mass of }KClO_3}{\text{Molar mass of }KClO_3}[/tex]
Molar mass of [tex]KClO_3[/tex] = 122.55 g/mole
[tex]\text{Moles of }KClO_3=\frac{7.5g}{122.55g/mole}=0.06119mole[/tex]
Now we have to calculate the moles of [tex]O_2[/tex] gas.
The balanced chemical reaction is:
[tex]2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)[/tex]
From the reaction we conclude that,
As, 2 moles of [tex]KClO_3[/tex] decomposes to give 3 moles of [tex]O_2[/tex]
So, 0.06119 moles of [tex]KClO_3[/tex] decomposes to give [tex]\frac{0.06119}{2}\times 3=0.0918[/tex] moles of [tex]O_2[/tex]
Now we have to calculate the volume of [tex]O_2[/tex] gas.
Using ideal gas equation :
[tex]PV=nRT[/tex]
where,
P = Pressure of [tex]O_2[/tex] gas = 1.00 atm
V = Volume of [tex]O_2[/tex] gas = ?
n = number of moles [tex]O_2[/tex] = 0.0918 mole
R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]
T = Temperature of [tex]O_2[/tex] gas = [tex]25^oC=273+25=298K[/tex]
Putting values in above equation, we get:
[tex]1.00atm\times V=0.0918mole\times (0.0821L.atm/mol.K)\times 298K[/tex]
[tex]V=2.24L[/tex]
Therefore, the volume of [tex]O_2[/tex] gas produced is 2.24 L