2KClO3 (s) ---> 2KCl (s) + 3O2 (g)

For every mole of KClO3 that decomposes in the above experiment, how many grams of oxygen gas are produced?

A.) 3 B.) 16 C.) 48

How many mols O2 are produced? That's 1 mol KClO3 x (3 mols O2/2 mol KClO3) = 3/2 mol.

3/2 mol x molar mass = g = about 48 or so. right?

To determine how many grams of oxygen gas are produced when one mole of KClO3 decomposes in the given reaction, we need to use the molar mass and stoichiometry of the reaction.

1. Find the molar mass of KClO3:
- K (potassium) has a molar mass of 39.1 g/mol
- Cl (chlorine) has a molar mass of 35.5 g/mol
- O (oxygen) has a molar mass of 16.0 g/mol
- Add up the molar masses of each element: 39.1 + 35.5 + (3 * 16.0) = 122.6 g/mol

2. Using the stoichiometry of the reaction, we see that for every 2 moles of KClO3, 3 moles of O2 are produced.
- This means that the mole ratio of moles of O2 to moles of KClO3 is 3:2.

3. Now, we can set up a proportion to find the moles of O2 produced:
- 3 moles of O2 / 2 moles of KClO3 = x moles of O2 / 1 mole of KClO3

- Cross-multiply: 2 * x = 3 * 1
- Simplify: 2x = 3
- Divide both sides by 2: x = 3/2

Therefore, for every mole of KClO3 that decomposes, 3/2 moles of O2 are produced.

4. Finally, to find the mass of O2 produced, we multiply the moles of O2 by the molar mass of O2:
- Mass of O2 = (3/2) moles of O2 * 32.0 g/mol (molar mass of O2)
- Mass of O2 = 48.0 g

Therefore, for every mole of KClO3 that decomposes, 48 grams of oxygen gas are produced. Hence, the correct answer is C.) 48.