Potassium superoxide,`KO_(2)`, is used in rebreathing gas masks to generate oxygen :
`" " KO_(2)(s) + H_(2)O(I) to KOH +O_(2)(g) `
If a reaction vessel contains 0.15 mol `KO_(2)` and 0.10 mol `H_(2)O` , how many moles of `O_(2)` can be produced?

Answer

To solve the problem of how many moles of oxygen (O₂) can be produced from the reaction of potassium superoxide (KO₂) and water (H₂O), we will follow these steps:

### Step 1: Write the balanced chemical equation
The unbalanced reaction is:
\[ \text{KO}_2 (s) + \text{H}_2\text{O} (l) \rightarrow \text{KOH} (s) + \text{O}_2 (g) \]

To balance this equation, we need to ensure that the number of atoms of each element is the same on both sides. The balanced equation is:
\[ 4 \text{KO}_2 (s) + 2 \text{H}_2\text{O} (l) \rightarrow 4 \text{KOH} (s) + 3 \text{O}_2 (g) \]

### Step 2: Identify the moles of reactants
From the problem, we have:
- Moles of KO₂ = 0.15 mol
- Moles of H₂O = 0.10 mol

### Step 3: Determine the limiting reagent
From the balanced equation, we see that:
- 4 moles of KO₂ react with 2 moles of H₂O.

Now, we can find out how many moles of H₂O are required for the available moles of KO₂:
- For 0.15 moles of KO₂:
\[ \text{Required moles of H}_2\text{O} = \frac{2 \text{ moles H}_2\text{O}}{4 \text{ moles KO}_2} \times 0.15 \text{ moles KO}_2 = 0.075 \text{ moles H}_2\text{O} \]

Since we have 0.10 moles of H₂O available, which is more than the required 0.075 moles, KO₂ is the limiting reagent.

### Step 4: Calculate the moles of O₂ produced
According to the balanced equation:
- 4 moles of KO₂ produce 3 moles of O₂.

Now, we can find out how many moles of O₂ are produced from 0.15 moles of KO₂:
\[ \text{Moles of O}_2 = \frac{3 \text{ moles O}_2}{4 \text{ moles KO}_2} \times 0.15 \text{ moles KO}_2 = 0.1125 \text{ moles O}_2 \]

### Final Answer
The number of moles of O₂ that can be produced is **0.1125 moles**.

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