Final answer:
About 29.9405 moles of KClO₃ are needed to produce 1006 L of O₂ at STP, calculated using stoichiometry and the known volume to moles conversion for gases.
Step-by-step explanation:
To calculate the number of moles of KClO₃ needed to produce 1006 L of O₂ at standard temperature and pressure (STP), we'll start with the equation 2KClO₃(s) → 2KCl(s) + 3O₂(g). This equation shows that 2 moles of KClO₃ produce 3 moles of O₂.
First, we need to convert the volume of O₂ to moles using the ideal gas law, which assumes that 1 mole of any gas at STP occupies 22.4 L. Therefore:
- 1006 L O₂ x (1 mole O₂ / 22.4 L O₂) = 44.9107 moles O₂
Next, we apply stoichiometry to determine the moles of KClO₃ needed:
- 44.9107 moles O₂ x (2 moles KClO₃ / 3 moles O₂) = 29.9405 moles KClO₃
Therefore, approximately 29.9405 moles of KClO₃ are required to produce 1006 L of O₂ at STP based on the given chemical reaction.