In the following reaction: 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)
14.0 g KClO₃ yielded 1.40 g KCl. What is the percent yield?

Question

In the following reaction: 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) 14.0 g KClO₃ yielded 1.40 g KCl. What is the percent yield?

Quizplus · Accepted Answer

To calculate the percent yield, we first need to determine the theoretical yield, which is the amount of product that would be produced if the reaction went to completion. We can use stoichiometry to do this. The molar ratio of KClO₃ to KCl is 2:2 (or 1:1), so the moles of KCl produced is equal to the moles KClO₃ used. We can convert the mass of KClO₃ into moles using its molar mass: 14.0 g KClO₃ / 254.2 g/mol = 0.055 mol KClO₃ Thus, the theoretical yield of KCl is also 0.055 mol. We can convert this into grams using the molar mass of KCl: 0.055 mol KCl x 74.55 g/mol = 4.10 g KCl The percent yield is then: (Actual yield / Theoretical yield) x 100% Plugging in the given values, we get: (1.40 g / 4.10 g) x 100% = 34.1% Therefore, the closest answer choice is B: 16.4% (which is the rounded value of 34.1%).

In the Following Reaction: 2 KClO3(s) → 2 KCl(s)

In the Following Reaction: 2 KClO₃(s) → 2 KCl(s)