Question

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.)

Answer 1

KClO3(s)+Δ→KCl(s)+32O2(g)

Approx. 3L of dioxygen gas will be evolved.

Explanation:

We assume that the reaction as written proceeds quantitatively.

Moles of KClO3(s) = 10.0⋅g122.55⋅g⋅mol−1 = 0.0816⋅mol

And thus 32×0.0816⋅mol dioxygen are produced, i.e. 0.122⋅mol.

At STP, an Ideal Gas occupies a volume of 22.4⋅L⋅mol−1.

And thus, volume of gas produced = 22.4⋅L⋅mol−1×0.0816⋅mol≅3L

Note that this reaction would not work well without catalysis, typically MnO2.

Answer 2

Answer:

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 * $\frac{1molKClO_{3}}{122.55gKClO_{3}} *\frac{3molO_{2}}{2molKClO_{3}} =$ 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