Calculate the theoretical volume yield ( In mL ) of oxygen gas produced at STP conditions, when 0.920 g of $$ \mathrm{KClO}_{3} $$ decomposes. Show all work and circle your correctly rounded answer. Report your answer to 3 sf .

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To calculate the theoretical volume yield of oxygen gas produced from the decomposition of potassium chlorate ($$ \mathrm{KClO}_{3} $$), we will follow these steps:

1. **Write the balanced chemical equation for the decomposition of $$ \mathrm{KClO}_{3} $$**:
   $$
   2 \, \mathrm{KClO}_{3} \, (s) \rightarrow 2 \, \mathrm{KCl} \, (s) + 3 \, \mathrm{O}_{2} \, (g)
   $$

2. **Calculate the molar mass of $$ \mathrm{KClO}_{3} $$**:
   - Potassium (K): 39.10 g/mol
   - Chlorine (Cl): 35.45 g/mol
   - Oxygen (O): 16.00 g/mol (3 O atoms)

$$
   \text{Molar mass of } \mathrm{KClO}_{3} = 39.10 + 35.45 + (3 \times 16.00) = 122.55 \, \text{g/mol}
   $$

3. **Calculate the number of moles of $$ \mathrm{KClO}_{3} $$** in 0.920 g:
   $$
   n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.920 \, \text{g}}{122.55 \, \text{g/mol}} \approx 0.00751 \, \text{mol}
   $$

4. **Use the stoichiometry of the reaction to find the moles of $$ \mathrm{O}_{2} $$** produced:
   From the balanced equation, 2 moles of $$ \mathrm{KClO}_{3} $$ produce 3 moles of $$ \mathrm{O}_{2} $$:
   $$
   \text{Moles of } \mathrm{O}_{2} = n \times \frac{3}{2} = 0.00751 \, \text{mol} \times \frac{3}{2} \approx 0.01127 \, \text{mol}
   $$

5. **Calculate the volume of $$ \mathrm{O}_{2} $$** at STP:
   At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 L. Therefore, the volume of $$ \mathrm{O}_{2} $$ produced is:
   $$
   V = n \times 22.4 \, \text{L/mol} = 0.01127 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 0.2525 \, \text{L}
   $$

6. **Convert the volume from liters to milliliters**:
   $$
   V = 0.2525 \, \text{L} \times 1000 \, \text{mL/L} = 252.5 \, \text{mL}
   $$

7. **Round the answer to 3 significant figures**:
   The final answer is:
   $$
   \boxed{253} \, \text{mL}
   $$

Thus, the theoretical volume yield of oxygen gas produced at STP conditions when 0.920 g of $$ \mathrm{KClO}_{3} $$ decomposes is 253 mL.
The theoretical volume of oxygen gas produced is 253 mL.

Calculate the theoretical volume yield ( In mL ) of oxygen gas produced at STP conditions, when 0.920 g of \( \mathrm{KClO}_{3} \) decomposes. Show all work and circle your correctly rounded answer. Report your answer to 3 sf .

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