O2 is evolved by heating KClO3 using MnO2 as a catalyst2KClO3- 2KCl-3O2Calculate the volume occupied by 0-01 mole of CO2 at STP-

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Standard X

Chemistry

Molecular Mass

O2 is evolved...

Question

$O_{2}$ is evolved by heating ${KClO}_{3}$ using ${MnO}_{2}$ as a catalyst
$2 {KClO}_{3} \to 2 KCl + 3 O_{2}$
Calculate the volume occupied by $0.01$ mole of ${CO}_{2}$ at STP.

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Solution

$1$ mole of carbon dioxide gas at S.T.P occupies $= 22.4 L$
$0.01$ mole of carbon dioxide gas at S.T.P occupies $= 22.4 \times 0.01 = 0.224 L$
Therefore, the volume occupied by $0.01$ mole of ${CO}_{2}$ at STP is $0.224 L$ .

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$O_{2}$ is evolved by heating ${KClO}_{3}$ using ${MnO}_{2}$ as a catalyst
$2 {KClO}_{3} \to 2 KCl + 3 O_{2}$

Calculate the number of moles of oxygen present in the above volume and also the number of molecules.

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O2 is evolved by heating KClO3 using MnO2 as a catalyst2KClO3→2KCl+3O2Calculate the volume occupied by 0.01 mole of CO2 at STP.

1 mole of carbon dioxide gas at S.T.P occupies =22.4L0.01 mole of carbon dioxide gas at S.T.P occupies =22.4×0.01=0.224LTherefore, the volume occupied by 0.01 mole of CO2 at STP is 0.224L.

$O_{2}$ is evolved by heating ${KClO}_{3}$ using ${MnO}_{2}$ as a catalyst
$2 {KClO}_{3} \to 2 KCl + 3 O_{2}$
Calculate the volume occupied by $0.01$ mole of ${CO}_{2}$ at STP.

$1$ mole of carbon dioxide gas at S.T.P occupies $= 22.4 L$
$0.01$ mole of carbon dioxide gas at S.T.P occupies $= 22.4 \times 0.01 = 0.224 L$
Therefore, the volume occupied by $0.01$ mole of ${CO}_{2}$ at STP is $0.224 L$ .