Equal masses of `KClO_(3)` undergoes different reaction in two different container:
(i) `2KClO_(3) to 2KCl + 3O_(2)`
(ii) `4KClO_(3)to KCl + 3KClO_(4)`
Mass ratio of the KCl produced in respective reaction is x : 1. Value of 'x' will be

Options

A. 4
B. 2
C. 0.25
D. 3

Answer

To solve the problem, we need to analyze the two reactions given and determine the mass ratio of KCl produced from equal masses of KClO3 undergoing these reactions.

### Step-by-Step Solution:

1. **Identify the Reactions:**
- Reaction (i): \( 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \)
- Reaction (ii): \( 4 \text{KClO}_3 \rightarrow \text{KCl} + 3 \text{KClO}_4 \)

2. **Let the Mass of KClO3:**
- Assume the mass of KClO3 used in both reactions is \( W \) grams.

3. **Calculate Moles of KClO3:**
- The molar mass of KClO3 is approximately \( 122.55 \, \text{g/mol} \).
- Number of moles of KClO3 = \( \frac{W}{122.55} \) moles.

4. **Calculate KCl Produced from Reaction (i):**
- From Reaction (i), \( 2 \) moles of KClO3 produce \( 2 \) moles of KCl.
- Therefore, \( 1 \) mole of KClO3 produces \( 1 \) mole of KCl.
- Moles of KCl produced from \( \frac{W}{122.55} \) moles of KClO3 = \( \frac{W}{122.55} \) moles of KCl.
- Mass of KCl produced = Moles × Molar mass of KCl.
- Molar mass of KCl is approximately \( 74.55 \, \text{g/mol} \).
- Mass of KCl from Reaction (i) = \( \frac{W}{122.55} \times 74.55 \).

5. **Calculate KCl Produced from Reaction (ii):**
- From Reaction (ii), \( 4 \) moles of KClO3 produce \( 1 \) mole of KCl.
- Therefore, \( 1 \) mole of KClO3 produces \( \frac{1}{4} \) mole of KCl.
- Moles of KCl produced from \( \frac{W}{122.55} \) moles of KClO3 = \( \frac{1}{4} \times \frac{W}{122.55} \) moles of KCl.
- Mass of KCl produced = Moles × Molar mass of KCl.
- Mass of KCl from Reaction (ii) = \( \frac{1}{4} \times \frac{W}{122.55} \times 74.55 \).

6. **Set Up the Mass Ratio:**
- The mass of KCl from Reaction (i) = \( \frac{W \times 74.55}{122.55} \).
- The mass of KCl from Reaction (ii) = \( \frac{1}{4} \times \frac{W \times 74.55}{122.55} \).
- The ratio of the masses of KCl produced from both reactions is:
\[ \text{Ratio} = \frac{\frac{W \times 74.55}{122.55}}{\frac{1}{4} \times \frac{W \times 74.55}{122.55}} = \frac{4}{1} \]

7. **Conclusion:**
- The mass ratio of KCl produced in the two reactions is \( x:1 \) where \( x = 4 \).

### Final Answer:
The value of \( x \) is \( 4 \).

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