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+6KOH⟶5KCl+KClO3+3H2O

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0.150 moles Cl2 x 1 mole KClO3/3 moles Cl2 x 122.5 g/1 mole KCIO3,

0.150 moles Cl2 x 1 mole KCIO3 /3 moles Cl3 x l mole KCIO3/122.5 g

0.150 moles Cl2 x 3 moles Cl2/1 mole KClO3 x 122.5 g/1 mole KClO2,

0.150 moles Cl2 x 3 moles Cl2/1 mole KCIO3, x 1 mole KCIO3./122.5 g

answer is A.

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Detailed Solution

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

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