How many moles of KClO3 are needed to form 2.8 L of O2, measured at STP, according to the following reaction: 2KClO3 🡪 2KCl + 3O2

Step 1: Identify the given information

• Volume of O2 = 2.8 L
• We are at Standard Temperature and Pressure (STP), which means that 1 mole of any gas occupies 22.4 L.

Step 2: Convert the volume of O2 to moles using the molar volume of a gas at STP

• Moles of O2 = Volume of O2 / Molar volume at STP = 2.8 L / 22.4 L/mol = 0.125 mol O2

Step 3: Use the stoichiometry of the reaction to find the moles of KClO3

• From the balanced chemical equation, we can see that 2 moles of KClO3 produce 3 moles of O2. Therefore, the ratio of KClO3 to O2 is 2:3.
• Moles of KClO3 = (Moles of O2 * Stoichiometric coefficient of KClO3) / Stoichiometric coefficient of O2 = (0.125 mol O2 * 2) / 3 = 0.0833 mol KClO3

So, 0.0833 moles of KClO3 are needed to form 2.8 L of O2 at STP.