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Which of the following setups is correct to calculate the weight (in g) of KCIO3 produced from the reaction of 0.150 moles of Cl2?
3Cl2+6KOH5KCl+KClO3+3H2O

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a
0.150 moles Cl2 x 1 mole KClO3/3 moles Cl2 x 122.5 g/1 mole KCIO3,
b
0.150 moles Cl2 x 1 mole KCIO3 /3 moles Cl3 x l mole KCIO3/122.5 g
c
0.150 moles Cl2 x 3 moles Cl2/1 mole KClO3 x 122.5 g/1 mole KClO2,
d
0.150 moles Cl2 x 3 moles Cl2/1 mole KCIO3, x 1 mole KCIO3./122.5 g

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detailed solution

Correct option is A

Factor label method. number of moles of  KClO3 produced =0.153 weight of KClO3 =0.153×122.5

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