Consider the following series of reaction:
`Cl_(2) + 2NaOH rarr NaCl + NaClO + H_(2)O`
`3NaClO rarr 2NaCl + NaClO_(3)`.
`4 NaClO_(3) rarr 3NaClO_(4) + NaCl`
How much `Cl_(2)` is needed to prepare `122.5 g NaClO_(4)` by above sequence?

Options

A. `284.0 g`
B. `213.0 g`
C. `142.0g`
D. `71.0 g`

Answer

A

Explanation

Factor label method: `(Mw "of" NaClO_(4) = 122.5 Mw "of" NaClO_(3) = 106.5 g mol^(-1))`
`x g` of `Cl_(2) = ((122.5 g NaClO_(4))/({:(122.5 g NaClO_(4)),("per mol of" NaClO_(4)):})) ((4 "mol"NaClO_(3))/(2 "mol" NaClO_(4)))`
`((3 "mol" NaClO)/(1 "mol" NaClO_(3)))((1 "mol"Cl_(2))/(1 "mol"NaClO))((71.0 g Cl_(2))/(1 "mol" Cl_(2)))`
`= (cancel(122.5))/(cancel(122.5)) xx (4)/(cancel(3)) xx (cancel(3))/(1) xx (1)/(1) xx (71.0)/(1) = 284.0 g`
Mole method:
`n(NaClO) = n(Cl_(2))`,
`n(NaClO_(3)) = (1)/(3) n(NaClO) = (1)/(3) (Cl_(2))`
`n(NaClO_(4)) = (3)/(4) n(NaClO_(3)) = (3)/(4) xx (1)/(3) n(Cl_(2)) = (1)/(4) n(Cl_(2))`
`n(NaClO_(4)) = (122.5 g NaClO_(4))/(122.5 "of" NaClO_(4)//"mole"NaClO_(4))`
`= 1.0 "mol" NaClO_(4)`
`n(Cl_(2)) = 4 xx 1.0 = 4 "mol" Cl_(2)`
Mass of `(Cl_(2)) = 4xx71.0g = 284 g Cl^(2)`