The interaction of `O_(3)` with potassium hydroxide gives ozonide according to the following equation.
`underset(168g)(3KOH(s))+2O_(3)(g) to underset(174g)(2KO_(3)(s))+KOH.H_(2)O(s)+(1)/(2)O_(2)(g)`
The ozonide `KO_(3)` slowly decomposes to `KO_(2)` and oxygen
`underset(174g)(2KO_(3)(s)) to underset(142g)(2KO_(2)(s))+O_(2)(g)`
The mass of `KO_(2)` produced by the reaction of 75.0g of KOH is

Options

A. 6.34 g
B. 63.4 g
C. 634 g
D. 0.634 g

Explanation

To determine the mass of `KO2` produced from the reaction of `75.0 g` of `KOH`, we can follow these steps:

### Step 1: Write down the balanced chemical equations
The first reaction is:
\[ 3 \text{KOH} + 2 \text{O}_3 \rightarrow 2 \text{KO}_3 + \text{KOH} \cdot \text{H}_2\text{O} + \frac{1}{2} \text{O}_2 \]
The second reaction is:
\[ 2 \text{KO}_3 \rightarrow 2 \text{KO}_2 + \text{O}_2 \]

### Step 2: Determine the molar masses
- Molar mass of `KOH` = 39 (K) + 16 (O) + 1 (H) = 56 g/mol
- Molar mass of `KO3` = 39 (K) + 48 (O3) = 87 g/mol
- Molar mass of `KO2` = 39 (K) + 32 (O2) = 71 g/mol

### Step 3: Calculate moles of `KOH`
Using the molar mass of `KOH`, we can calculate the number of moles in `75.0 g` of `KOH`:
\[ \text{Moles of KOH} = \frac{\text{mass}}{\text{molar mass}} = \frac{75.0 \text{ g}}{56 \text{ g/mol}} \approx 1.339 \text{ moles} \]

### Step 4: Use stoichiometry to find moles of `KO3` produced
From the balanced equation, `3 moles of KOH` produce `2 moles of KO3`. Therefore, we can set up a ratio:
\[ \text{Moles of KO3} = \left(\frac{2 \text{ moles KO3}}{3 \text{ moles KOH}}\right) \times 1.339 \text{ moles KOH} \approx 0.892 \text{ moles KO3} \]

### Step 5: Calculate moles of `KO2` produced
From the second reaction, `2 moles of KO3` produce `2 moles of KO2`. Thus,
\[ \text{Moles of KO2} = \left(\frac{2 \text{ moles KO2}}{2 \text{ moles KO3}}\right) \times 0.892 \text{ moles KO3} = 0.892 \text{ moles KO2} \]

### Step 6: Calculate the mass of `KO2`
Now, we can find the mass of `KO2` produced using its molar mass:
\[ \text{Mass of KO2} = \text{moles} \times \text{molar mass} = 0.892 \text{ moles} \times 71 \text{ g/mol} \approx 63.39 \text{ g} \]

### Final Answer
The mass of `KO2` produced by the reaction of `75.0 g` of `KOH` is approximately **63.39 g**.

---