Potassium superoxide ,`KO_(2)`, is utilised in closed system breathing apparatus. Exhaled air contains `CO_(2)` and `H_(2)O` ,both of which are removed and the removal of water generates oxygen for breathing by the reaction
`4KO_(2)`(s) + `2H_(2)O(l) to 3O_(2)(g) + 4KOH(s)`
The potassium hydroxide removes `CO_(2)` from the apparatus by the reaction :
`KOH(s) + CO_(2)(g) to KHCO_(3)`(s)
(a) What mass of `KO_(2)` generates 24 gm of oxygen ?
(b) What mass of `CO_(2)` can be removed from from the apparatus by 100 gm of `KO_(2)` ?

Answer

To solve the given problem, we will break it down into two parts as specified in the question.

### Part (a): What mass of `KO_(2)` generates 24 g of oxygen?

1. **Write the balanced chemical equation**:
\[ 4 \text{KO}_2 (s) + 2 \text{H}_2\text{O} (l) \rightarrow 3 \text{O}_2 (g) + 4 \text{KOH} (s) \]

2. **Determine the mole ratio**:
From the equation, we see that 4 moles of `KO_(2)` produce 3 moles of `O_(2)`. Therefore, the ratio can be expressed as:
\[ \frac{4 \text{ moles KO}_2}{3 \text{ moles O}_2} \]

3. **Calculate moles of `O_(2)`**:
The molar mass of oxygen (`O_(2)`) is 32 g/mol. To find the number of moles of `O_(2)` in 24 g:
\[ \text{Moles of } O_2 = \frac{24 \text{ g}}{32 \text{ g/mol}} = 0.75 \text{ moles} \]

4. **Use the mole ratio to find moles of `KO_(2)`**:
Using the mole ratio from step 2:
\[ \text{Moles of KO}_2 = 0.75 \text{ moles O}_2 \times \frac{4 \text{ moles KO}_2}{3 \text{ moles O}_2} = 1 \text{ mole KO}_2 \]

5. **Calculate the mass of `KO_(2)`**:
The molar mass of `KO_(2)` is 71 g/mol. Therefore, the mass of `KO_(2)` needed is:
\[ \text{Mass of KO}_2 = 1 \text{ mole} \times 71 \text{ g/mol} = 71 \text{ g} \]

**Final Answer for Part (a)**:
The mass of `KO_(2)` that generates 24 g of oxygen is **71 g**.

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### Part (b): What mass of `CO_(2)` can be removed from the apparatus by 100 g of `KO_(2)`?

1. **Write the relevant reaction**:
The reaction for the removal of `CO_(2)` by `KOH` is:
\[ \text{KOH} (s) + \text{CO}_2 (g) \rightarrow \text{KHCO}_3 (s) \]

2. **Determine moles of `KO_(2)`**:
First, calculate the number of moles of `KO_(2)` in 100 g:
\[ \text{Moles of KO}_2 = \frac{100 \text{ g}}{71 \text{ g/mol}} \approx 1.408 \text{ moles} \]

3. **Determine moles of `KOH` produced**:
From the balanced equation, 4 moles of `KO_(2)` produce 4 moles of `KOH`. Thus:
\[ \text{Moles of KOH} = 1.408 \text{ moles KO}_2 \approx 1.408 \text{ moles KOH} \]

4. **Use the mole ratio for `KOH` and `CO_(2)`**:
From the reaction, 1 mole of `KOH` reacts with 1 mole of `CO_(2)`. Therefore:
\[ \text{Moles of CO}_2 = 1.408 \text{ moles KOH} \approx 1.408 \text{ moles CO}_2 \]

5. **Calculate the mass of `CO_(2)`**:
The molar mass of `CO_(2)` is 44 g/mol. Thus, the mass of `CO_(2)` that can be removed is:
\[ \text{Mass of CO}_2 = 1.408 \text{ moles} \times 44 \text{ g/mol} \approx 61.792 \text{ g} \approx 61.6 \text{ g} \]

**Final Answer for Part (b)**:
The mass of `CO_(2)` that can be removed from the apparatus by 100 g of `KO_(2)` is approximately **61.6 g**.

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