Final answer:
The decomposition of potassium chlorate that produced 24.0 g of KCl also produced approximately 16.0 g of O₂, corresponding to option (a) 16.0 g O₂.
Step-by-step explanation:
When potassium chlorate (KCIO₃) is heated, it decomposes into potassium chloride (KCl) and oxygen gas (O₂). The balanced equation for this reaction is 2 KCIO₃ → 2 KCl + 3 O₂. If 24.0 g of KCl is produced, we can use stoichiometry to find out how many grams of O₂ were produced.
Firstly, we need to find the molar mass of KCIO₃ and KCl to convert grams into moles:
- Molar mass of KCIO₃ = 39.1 (K) + 35.5 (Cl) + 3(16.0) (O) = 122.6 g/mol
- Molar mass of KCl = 39.1 (K) + 35.5 (Cl) = 74.6 g/mol
Next, we find the moles of KCl produced:
24.0 g KCl × (1 mol KCl / 74.6 g KCl) = 0.3217 mol KCl
Using the stoichiometry from the balanced equation, we can determine the moles of O₂:
2 mol KCl : 3 mol O₂
So,
0.3217 mol KCl × (3 mol O₂ / 2 mol KCl) = 0.48255 mol O₂
Finally, we convert moles of O₂ to grams:
0.48255 mol O₂ × 32.0 g/mol (molar mass of O₂) = 15.4416 g O₂
Therefore, if this reaction produced 24.0 g KCl, it also produced approximately 16.0 g of O₂, which corresponds to option (a) 16.0 g O₂.