Question

Calculate the volume of oxygen at NTP obtained by decomposing 12.26g of KCLO3(at wt. K=39.1, Cl=35.5 and O = 16)​

Answer 1

Answer: 3.61 L

Step-by-step explanation:

To calculate the moles, we use the equation:

`moles=\frac{\text {given mass}}{\text {Molar mass}}`

`moles=(12.26g)/(122.6g/mol)=0.1moles`

`2KClO_3\rightarrow 2KCl+3O_2`

2 moles of
`KClO_3` produce = 3 moles of
`O_2`

0.1 moles of
`KClO_3` produce =
`(3)/(2)* 0.1=0.15` moles of
`O_2`

According to the ideal gas equation:'

`PV=nRT`

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

`1* V=0.15* 0.0821* 293`

`V=3.61L`

Thus volume of oxygen at NTP obtained by decomposing 12.26 g of
`KClO_3` is 3.61 L