What volume of oxygen gas is released at stp if 10.0 g of potassium chlorate is decomposed? (the molar mass of kclo3 is 122.55 g/mol.)

2.74 L

Explanation:

The reaction that takes place is:

2KClO₃(s) + Δ → 2KCl (s) + 3O₂ (g)

To calculate the volume of oxygen gas released, we calculate the moles of O₂ produced, using the mass of reactant given by the problem:

10.0 g KClO₃ * 0.1224 mol O₂

At STP, 1 mol of gas occupies 22.4 L, so the amount occupied by 0.1224 mol is:

0.1224 mol * 22.4 L·mol⁻¹= 2.74 L