15 g sample of `BaO_(2)` is heated to `794^(@)C` in a closed evacuated vessel of 5 litre capacity. How many g of peroxide are converted to `BoO(s)`?
`2BaO_(2)(s)hArr2BaO(s)+O_(2)(g), K_(rho)=0.5` atm

Answer

To solve the problem step by step, we will follow the reaction and use the given data to find out how many grams of barium peroxide (BaO₂) are converted to barium oxide (BaO).

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction given is:
\[ 2 \text{BaO}_2 (s) \rightleftharpoons 2 \text{BaO} (s) + \text{O}_2 (g) \]
The equilibrium constant \( K_p \) is given as 0.5 atm.

2. **Calculate the Moles of Oxygen at Equilibrium**:
From the equilibrium constant expression, we know that:
\[ K_p = \frac{P_{\text{O}_2}}{1} = 0.5 \text{ atm} \]
This means the partial pressure of oxygen at equilibrium is 0.5 atm.

3. **Use the Ideal Gas Law to Find Moles of Oxygen**:
We can use the ideal gas law \( PV = nRT \) to find the number of moles of oxygen (\( n \)):
- \( P = 0.5 \) atm (partial pressure of \( O_2 \))
- \( V = 5 \) L (volume of the vessel)
- \( R = 0.0821 \) L·atm/(K·mol) (ideal gas constant)
- \( T = 794 + 273 = 1067 \) K (temperature in Kelvin)

Rearranging the ideal gas law to solve for \( n \):
\[ n = \frac{PV}{RT} = \frac{0.5 \times 5}{0.0821 \times 1067} \]

Calculating \( n \):
\[ n = \frac{2.5}{87.67} \approx 0.0285 \text{ moles of } O_2 \]

4. **Relate Moles of Oxygen to Moles of Barium Peroxide**:
According to the stoichiometry of the reaction, 2 moles of \( \text{BaO}_2 \) produce 1 mole of \( O_2 \). Therefore, the moles of \( \text{BaO}_2 \) that decomposed can be calculated as:
\[ n(\text{BaO}_2) = 2 \times n(O_2) = 2 \times 0.0285 \approx 0.057 \text{ moles of } \text{BaO}_2 \]

5. **Calculate the Mass of Barium Peroxide Decomposed**:
The molar mass of \( \text{BaO}_2 \) is:
- Ba: 137.33 g/mol
- O: 16.00 g/mol (2 O atoms)
\[ \text{Molar mass of } \text{BaO}_2 = 137.33 + (2 \times 16.00) = 169.33 \text{ g/mol} \]

Now, we can find the mass of \( \text{BaO}_2 \) that decomposed:
\[ \text{Mass} = n \times \text{Molar mass} = 0.057 \times 169.33 \approx 9.65 \text{ g} \]

### Final Answer:
Approximately **9.65 g** of barium peroxide are converted to barium oxide.