Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO 3 (s). The equation for the reaction is 2 KClO 3 ⟶ 2 KCl + 3 O 2 Calculate how many grams of O 2 (g) can be produced from heating 34.4 g KClO 3 (s).

Molar mass of KClO3 equals:

M (KCIO3) = M (K) + M (CI) + 3M (O) = 39.1 + 35.5 + 3*16.0 = 122.6 g

Mass of 2 moles of potassium chlorate equals:

2 * 122.6 = 245.2 g

Molar mass of O2 equals:

M (O2) = 2M (O) = 2 * 16.0 = 32.0 g

Mass of 3 moles of O2 equals:

3 ∙ 32.0 = 96 g

Therefore:

If 245.2 g of KClO3 produce 96.0 g of O2

34.4 g of KClO3 - x g of O

Hence x = (34.4 * 96.0) : 245.2

x = 13.47g

Grams of oxygen produced = 13.47 grams

Explanation:

Firstly, let's balance the equation of the chemical reaction. Heating potassium trioxochlorate produces potassium chloride and oxygen molecules.

KClO 3 → KCl + O2

Balance the equation

2KClO3 → 2KCl + 3O2

From the chemical equation 2 moles of KClO3 decompose to produce 2 moles of KCl and 3 moles of Oxygen molecule.

atomic mass of K = 39.1g

atomic mass of Cl = 35.5g

atomic mass of O = 16g

The molecular mass of KClO3 = 39.1 + 35.5 + 16 * 3 = 122.6 g/mol

The mass of 2 moles of KClO3 = 2 * 122.6 = 245.2 g

molecular mass of O2 = 32 g

Mass of 3 moles of oxygen molecule = 3 * 32 = 96 grams

Using proportion

245.2g of KClO3 produces 96 g of oxygen

34.4 g will produce? grams of Oxygen

Grams of oxygen produced = 34.4 * 96/245.2

Grams of oxygen produced = 3302.4 / 245.2

Grams of oxygen produced = 13.4681892333

Grams of oxygen produced = 13.47 grams