A 1.0 g sample of `KCIO_3` was heated under such conditions that a part of it decomposed according to the equation:
(i) `2KCIO_3 to 2KCI + 3O_2` and the remaining underwent a change according to the equation
(ii) `4KCIO_3 to 3KClO_4 + KCI`.
If the amount of oxygen evolved was 146.8 mL at S.T.R, calculate the percentage by weight of `KClO_4` in the residue.

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Calculate the moles of oxygen evolved
Given the volume of oxygen evolved is 146.8 mL at STP. We need to convert this volume into liters:

\[ \text{Volume of } O_2 = 146.8 \, \text{mL} = 0.1468 \, \text{L} \]

Using the ideal gas law, at STP, 1 mole of gas occupies 22.4 L. Thus, we can calculate the moles of oxygen:

\[ \text{Moles of } O_2 = \frac{\text{Volume of } O_2}{22.4} = \frac{0.1468}{22.4} \approx 0.00655 \, \text{moles} \]

### Step 2: Calculate the moles of KClO3 decomposed in the first reaction
From the first reaction:

\[ 2 \, KClO_3 \rightarrow 2 \, KCl + 3 \, O_2 \]

From the stoichiometry, 3 moles of \( O_2 \) are produced from 2 moles of \( KClO_3 \). Therefore, the moles of \( KClO_3 \) that decomposed can be calculated as follows:

\[ \text{Moles of } KClO_3 = \frac{2}{3} \times \text{Moles of } O_2 = \frac{2}{3} \times 0.00655 \approx 0.00436 \, \text{moles} \]

### Step 3: Calculate the mass of KClO3 used in the first reaction
The molar mass of \( KClO_3 \) is approximately 122.5 g/mol. Therefore, the mass of \( KClO_3 \) that decomposed in the first reaction is:

\[ \text{Mass of } KClO_3 = \text{Moles} \times \text{Molar Mass} = 0.00436 \times 122.5 \approx 0.534 \, \text{g} \]

### Step 4: Calculate the mass of KClO3 used in the second reaction
The total mass of the sample is 1.0 g. Thus, the mass of \( KClO_3 \) that underwent the second reaction is:

\[ \text{Mass of } KClO_3 \text{ in second reaction} = 1.0 \, \text{g} - 0.534 \, \text{g} \approx 0.466 \, \text{g} \]

### Step 5: Calculate the moles of KClO3 used in the second reaction
Using the molar mass of \( KClO_3 \):

\[ \text{Moles of } KClO_3 = \frac{0.466}{122.5} \approx 0.00380 \, \text{moles} \]

### Step 6: Calculate the moles of KClO4 produced
From the second reaction:

\[ 4 \, KClO_3 \rightarrow 3 \, KClO_4 + KCl \]

From the stoichiometry, 4 moles of \( KClO_3 \) produce 3 moles of \( KClO_4 \). Therefore, the moles of \( KClO_4 \) produced are:

\[ \text{Moles of } KClO_4 = \frac{3}{4} \times \text{Moles of } KClO_3 = \frac{3}{4} \times 0.00380 \approx 0.00285 \, \text{moles} \]

### Step 7: Calculate the mass of KClO4 produced
Using the molar mass of \( KClO_4 \) (approximately 138.5 g/mol):

\[ \text{Mass of } KClO_4 = \text{Moles} \times \text{Molar Mass} = 0.00285 \times 138.5 \approx 0.395 \, \text{g} \]

### Step 8: Calculate the moles of KCl produced
From the second reaction, the moles of \( KCl \) produced are:

\[ \text{Moles of } KCl = \frac{1}{4} \times \text{Moles of } KClO_3 = \frac{1}{4} \times 0.00380 \approx 0.00095 \, \text{moles} \]

### Step 9: Calculate the mass of KCl produced
Using the molar mass of \( KCl \) (approximately 74.5 g/mol):

\[ \text{Mass of } KCl = \text{Moles} \times \text{Molar Mass} = 0.00095 \times 74.5 \approx 0.071 \, \text{g} \]

### Step 10: Calculate the total mass of the residue
The total mass of the residue consists of \( KClO_4 \) and \( KCl \) produced from both reactions:

\[ \text{Total mass of residue} = \text{Mass of } KClO_4 + \text{Mass of } KCl = 0.395 \, \text{g} + 0.071 \, \text{g} \approx 0.466 \, \text{g} \]

### Step 11: Calculate the percentage by weight of KClO4 in the residue
Finally, we can calculate the percentage of \( KClO_4 \) in the residue:

\[ \text{Percentage of } KClO_4 = \left( \frac{\text{Mass of } KClO_4}{\text{Total mass of residue}} \right) \times 100 = \left( \frac{0.395}{0.466} \right) \times 100 \approx 84.7\% \]

### Final Answer
The percentage by weight of \( KClO_4 \) in the residue is approximately **84.7%**.