MnO2+2KClO3⟶K2MnO4+Cl2+2O2 Cl2+KMnO4⟶2KCl+MnO2+O2 Each reaction takes place to the extent of 50%. 11.2 L of O2 at STP is obtained from KClO3 using :
A
4 mol
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B
1.12 mol
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C
1.33 mol
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D
2.66 mol
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Solution
The correct option is A4 mol MnO2+2KClO3⟶K2MnO4+Cl2+2O2 Cl2+KMnO4⟶2KCl+MnO2+O2 According to the balanced reaction, 11.2 L O2 (0.5 mole) = 1 mole Cl2=4 moles KClO3 50% yield 50% yield