When potassium chlorate, KCI03 is decomposed, oxygen gas and potassium chloride are produced. 2 KCI 03 - 3 02 + 2 KCI What volume of oxygen gas is produced when 0.712 g of KCI03 decomposes at STP?
To solve this problem, we need to first calculate the number of moles of KClO3 that decompose, and then use the stoichiometry of the reaction to determine the volume of oxygen gas produced.
1. Calculate the number of moles of KClO3:
Molar mass of KClO3 = 39.10 g/mol (K) + 35.45 g/mol (Cl) + 3(16.00 g/mol) (O) = 122.55 g/mol
Number of moles of KClO3 = 0.712 g / 122.55 g/mol = 0.00581 mol
2. Use the stoichiometry of the reaction to determine the volume of oxygen gas produced:
From the balanced chemical equation, we see that 2 moles of KClO3 produce 3 moles of O2.
So, 0.00581 mol of KClO3 will produce (3/2) * 0.00581 mol = 0.00871 mol of O2
Now we use the ideal gas law: PV = nRT
At STP (Standard Temperature and Pressure), temperature (T) = 273 K and pressure (P) = 1 atm.
The volume of 1 mol of gas at STP is 22.4 L.
So, V = 0.00871 mol * 22.4 L/mol = 0.195 L
Therefore, the volume of oxygen gas produced when 0.712 g of KClO3 decomposes at STP is 0.195 L.