Answer :
Answer:
6.09 grams** of KCl would be produced from heating 10 grams of KClO3 under theoretical conditions.
Explanation:
To determine the mass of KCl produced in the reaction where potassium chlorate (KClO3) is heated to produce potassium chloride (KCl), we need to consider the stoichiometry of the reaction and the conservation of mass principle.
The balanced chemical equation for the decomposition of potassium chlorate is:
\[ 2 KClO3 \rightarrow 2 KCl + 3 O2 \]
This equation tells us that for every 2 moles of KClO3 decomposed, we get 2 moles of KCl produced.
Now, let's assume that during the lab activity, the technician heated a certain mass of KClO3 and obtained a certain mass of KCl. To find the mass of KCl produced, we need to know the initial mass of KClO3 used and the percent yield of the reaction (if provided). However, since the percent yield is not given, we will assume a theoretical yield based on stoichiometry.
Let's proceed with an example calculation:
Assume the technician used 10 grams of KClO3.
1. **Calculate the molar mass of KClO3:**
 - KClO3: Potassium (K) has a molar mass of approximately 39.1 g/mol, Chlorine (Cl) has a molar mass of approximately 35.5 g/mol, and Oxygen (O) has a molar mass of approximately 16.0 g/mol.
 Molar mass of KClO3 = 39.1 + 35.5 + (3 × 16.0) = 122.5 g/mol
2. **Calculate the number of moles of KClO3 used:**
 Moles of KClO3 = Mass / Molar mass = 10 g / 122.5 g/mol ≈ 0.0816 mol
3. **According to the balanced equation, the mole ratio of KClO3 to KCl is 1:1. Therefore, moles of KCl produced = moles of KClO3 used.**
 Moles of KCl produced = 0.0816 mol
4. **Calculate the mass of KCl produced:**
 Mass of KCl produced = Moles × Molar mass of KCl
 - Molar mass of KCl = 39.1 + 35.5 = 74.6 g/mol
 Mass of KCl produced = 0.0816 mol × 74.6 g/mol ≈ 6.09 grams
Therefore, based on these calculations, approximately **6.09 grams** of KCl would be produced from heating 10 grams of KClO3 under theoretical conditions. Actual lab results could vary based on the experimental conditions and percent yield achieved in the reaction.
