Final answer:
The theoretical yield of O2 from the decomposition of 10 g of KClO3 is 3.92 g, and the actual yield collected was 3.41 g. The percent yield is calculated as 87%.
Step-by-step explanation:
In a small-scale experiment, 10 g of KClO₃ was decomposed to collect 3.41 g of O₂ gas. To calculate the theoretical yield of O₂, stoichiometry must be applied:
- First, determine the moles of KClO₃ using its molar mass (1 mol KClO₃ = 122.55 g/mol).
- Next, use the balanced chemical equation to find the mol ratio of KClO₃ to O₂: 2KClO₃ (s) → 2KCl (s) + 3O₂ (g).
- Finally, convert the moles of O₂ to grams using the molar mass of O₂ (1 mol O₂ = 32.00 g/mol)
For 10 g of KClO₃, the theoretical yield of O₂ can be calculated as follows:
- 10 g KClO₃ ÷ 122.55 g/mol = 0.0816 mol KClO₃
- According to the balanced equation, 3 mol of O₂ are produced for every 2 moles of KClO₃, leading to 0.0816 mol KClO₃ x (3 mol O₂ / 2 mol KClO₃) = 0.1224 mol O₂
- 0.1224 mol O₂ x 32.00 g/mol = 3.92 g O₂
The theoretical yield of O₂ is 3.92 g. The actual yield given is 3.41 g. The percent yield can be calculated by dividing the actual yield by the theoretical yield and multiplying by 100:
Percent yield = (3.41 g / 3.92 g) x 100 = 87%.
Therefore, the theoretical yield is 3.92 g, the actual yield is 3.41 g, and the percent yield is 87%.