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How much Cl(2) in needed to prepare 106....

How much `Cl_(2)` in needed to prepare `106.5g`of `NaClO_(3)` by the above sequence?

A

`284.0 g`

B

`213.0 g`

C

`142.0g`

D

`71.0 g`

Text Solution

Verified by Experts

The correct Answer is:
B
`x g Cl_(2) = ((106.5 g NaClO_(3))/({:(106.5 g NaClO_(3)),("per mol"NaClO_(3)):}))((3 "mol" NaClO)/(1 "mol" NaClO_(3)))`
`((1 "mole" Cl_(2))/(1 "mol"NaClO))((71.0 g Cl_(2))/(1 "mol"Cl_(2)))`
`= (106.5)/(106.5) xx (3)/(1) xx (1)/(1) xx (17.0)/(1) = 213.0 g Cl_(2)`
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