`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)

Options

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`

Explanation

To solve the problem of preparing KClO₄ from the given reactions, we will analyze each reaction step-by-step and determine the required quantities using stoichiometric relationships.

### Step 1: Analyze the Reactions
We have three reactions provided:
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. \( 4 \text{KClO}_3 \rightarrow 3 \text{KCl} + \text{KClO}_4 \)

### Step 2: Determine Molar Relationships
From the reactions, we can derive the following stoichiometric relationships:
- From Reaction 1: 1 mole of Cl₂ produces 1 mole of KClO.
- From Reaction 2: 3 moles of KClO produce 1 mole of KClO₃.
- From Reaction 3: 4 moles of KClO₃ produce 1 mole of KClO₄.

### Step 3: Calculate Moles of KClO₄
To find out how many moles of KClO₄ can be produced from a certain amount of KClO, we can set up the following relationships:
- From Reaction 2: \( \text{Moles of KClO} = 3 \times \text{Moles of KClO}_3 \)
- From Reaction 3: \( \text{Moles of KClO}_3 = 4 \times \text{Moles of KClO}_4 \)

Combining these, we find:
\[ \text{Moles of KClO} = 3 \times (4 \times \text{Moles of KClO}_4) = 12 \times \text{Moles of KClO}_4 \]

### Step 4: Calculate Required Moles of Cl₂
From Reaction 1, since 1 mole of Cl₂ produces 1 mole of KClO, we can say:
\[ \text{Moles of Cl}_2 = \frac{1}{2} \times \text{Moles of KClO} \]

Substituting the expression for KClO in terms of KClO₄:
\[ \text{Moles of Cl}_2 = \frac{1}{2} \times 12 \times \text{Moles of KClO}_4 = 6 \times \text{Moles of KClO}_4 \]

### Step 5: Calculate Mass of Cl₂ Required
To find the mass of Cl₂ required, we will use the molar mass of Cl₂, which is \( 2 \times 35.5 = 71 \, \text{g/mol} \).
If we denote the moles of KClO₄ as \( n \), the mass of Cl₂ required is:
\[ \text{Mass of Cl}_2 = 6n \times 71 \]

### Step 6: Calculate Molar Mass of KClO₄
The molar mass of KClO₄ can be calculated as follows:
- K: 39 g/mol
- Cl: 35.5 g/mol
- O: \( 4 \times 16 = 64 \, \text{g/mol} \)

Thus, the molar mass of KClO₄ is:
\[ 39 + 35.5 + 64 = 138.5 \, \text{g/mol} \]

### Step 7: Calculate Moles of KClO₄ from Given Mass
Given that we have 277 grams of KClO₄:
\[ \text{Moles of KClO}_4 = \frac{277}{138.5} \approx 2 \, \text{moles} \]

### Step 8: Calculate Mass of Cl₂ Required
Using the moles of KClO₄ calculated:
\[ \text{Mass of Cl}_2 = 6 \times 2 \times 71 = 852 \, \text{grams} \]

### Conclusion
The mass of Cl₂ required to prepare 277 grams of KClO₄ is 852 grams.