Question

In the lab, 10.0 g of KClO₃ were carefully decomposed and 3.41 g of O₂ were collected. What is the percent yield?

Answer 1

$2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$

Find theoretical mass of O2:

$10.0 g (KClO_3) \times \frac{1 mole (KClO_3)}{122.55 g (KClO_3)}\times \frac{3 mole (O_2)}{2 mole (KClO_3)} \times \frac{32 g (O_2)}{1 mole (O_2)}= 3.92 g (O_2)$

Find percent yield:

percent yield % =$\frac {3.41 g(O_2)}{3.92 g(O_2)} \times 100\%=87\%$