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From the following reaction sequence C...

From the following reaction sequence
`Cl_2 + 2KOH to KCl + KClO + H_2O`
`3KClO to 2KCl + KClO_3`
`4KClO_3 to 3KClO_4 + KCl`
Calculate the mass of chlorine needed to produce 100 g of `KClO_4`

Text Solution

AI Generated Solution

To calculate the mass of chlorine needed to produce 100 g of KClO₄, we will follow these steps: ### Step 1: Write the overall reaction We have three reactions given: 1. \( \text{Cl}_2 + 2 \text{KOH} \rightarrow \text{KCl} + \text{KClO} + \text{H}_2\text{O} \) 2. \( 3 \text{KClO} \rightarrow 2 \text{KCl} + \text{KClO}_3 \) 3. \( 4 \text{KClO}_3 \rightarrow 3 \text{KClO}_4 + \text{KCl} \) ...
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Explore conceptually related problems

From the following series of reactions, {:(Cl_(2)+2KOH to KCl+KClO+H_(2)O),(" " 2KClO to 2KCl+KClO_(3)),(" "4KClO_(3) to 3 KClO_(4) + KCl):} calculate the mass of chlorine needed to produce 100 g of KCiO_(4)

Cl_(2) + KOH = KClO_(3) + KCl + H_(2) O

Knowledge Check

  • KClO_(4) can be prepared by following reactions: i. Cl_(2) + 2 KOH rarr KCl + KClO + H_(2) O ii. 3KClO rarr 2KCl + KClO_(3) iii. 4KClO_(3) rarr 3KlI_(4) + KCl (Atomic weight of K, Cl , and O are 369,35.5 and 16)

    A
    The amount of `Cl_(2)` required to prepare `277 g` of `KClO_(4)` by above series of reaction is `568 g`.
    B
    The volume of `KOH` in litres used by `Cl_(2)`, if `KOH` is `1.5 M`, is `1.067 L`.
    C
    The amount of `Cl_(2)` required to prepare `200 g "of" KClO_(4)` by above series of reactionis `284 g`
    D
    The volume of `KOH` in litres used by `Cl_(2)`, if `KOH` is `1.5 M`, is `10.76 L`
  • In the following reaction, calculate the mass of Cl_(2) required to produce 1385 g of KCiO_(4) (K=39):- Cl_(2)+ 2KOH to KCl +KClO +H_(2)O 3KClO to 2 KCl +KClO_(3) 4KClOto 3KClO_(4)+ KCl

    A
    40 g
    B
    2840 g
    C
    1420 g
    D
    710 g
  • In the following disproportionation of Cl_(2) in basic medium Cl_(2)+2KOH rarr KCl+KClO+H_(2)O Equivalent mass of Cl_(2) is

    A
    `35.50`
    B
    `71.00`
    C
    `47.33`
    D
    `11.83`
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