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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)​

Sagot :

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

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