Answer :
Answer: 3.61 L
Explanation:
To calculate the moles, we use the equation:
[tex]moles=\frac{\text {given mass}}{\text {Molar mass}}[/tex]
[tex]moles=\frac{12.26g}{122.6g/mol}=0.1moles[/tex]
[tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]
2 moles of [tex]KClO_3[/tex] produce = 3 moles of [tex]O_2[/tex]
0.1 moles of [tex]KClO_3[/tex] produce = [tex]\frac{3}{2}\times 0.1=0.15[/tex] moles of [tex]O_2[/tex]
According to the ideal gas equation:'
[tex]PV=nRT[/tex]
P = Pressure of the gas = 1 atm (NTP)
V= Volume of the gas = ?
T= Temperature of the gas = 20°C = (20+273) K = 293 K (NTP)
R= Value of gas constant in in kilopascals = 0.0821 Latm/K mol
[tex]1\times V=0.15\times 0.0821\times 293[/tex]
[tex]V=3.61L[/tex]
Thus volume of oxygen at NTP obtained by decomposing 12.26 g of [tex]KClO_3[/tex] is 3.61 L
