Question

O2IS EVOLVED BY HEATING KCL03 USING MN02 AS CATALYST AS 2KCLO3= 2KCL +302 FIND THE MASS OF OXYGEN RELEASED AND LOSS OF MASS
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Answer 1

Answer:

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2KCI03 Molecular mass of 2KCi03

= 2 x (39+35.5 +3x16) = 2 x 122.5 =245 g 2KCI = 2 (39 +35.5) =149 g 302 3 x 22.4 = 67.2 1 67.2 litre of Oz produced by KCI03 =245 g 1 litre of O2 produced byKCIo3 2KCI +302 245 67.2 0 67.2 6.72 of O2 produced by KCIO3 = 24.5

g i) 245 g of KCIo3 contains 3 moles of oxygen 24.5 g of KCIO3 contains = 3X24.5 245 0.3 mole since, 1 mole of oxygen contain = 6.022 x10° molecules therefore, 0.3 mole of oxygen contain = 6.022 x104 x0.3 = 1.8066 x 104 molecules

ii) 1 mole of CO2 Occupied = 22.41 0.01 mole of CO2 occupied 22.4 x 0.01 = 0.2241 7

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Answer 2

Explanation:

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2KCI03 Molecular mass of 2KCi03

= 2 x (39+35.5 +3x16) = 2 x 122.5 =245 g 2KCI = 2 (39 +35.5) =149 g 302 3 x 22.4 = 67.2 1 67.2 litre of Oz produced by KCI03 =245 g 1 litre of O2 produced byKCIo3 2KCI +302 245 67.2 0 67.2 6.72 of O2 produced by KCIO3 = 24.5

g i) 245 g of KCIo3 contains 3 moles of oxygen 24.5 g of KCIO3 contains = 3X24.5 245 0.3 mole since, 1 mole of oxygen contain = 6.022 x10° molecules therefore, 0.3 mole of oxygen contain = 6.022 x104 x0.3 = 1.8066 x 104 molecules

ii) 1 mole of CO2 Occupied = 22.41 0.01 mole of CO2 occupied 22.4 x 0.01 = 0.2241 7

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