Question

Refer to this equation: 3Cl2 + 6KOH --> KClO3 + 5KCl + 3H2O 30.0 grams of Cl2 is reacted with excess KOH. If 34.5 g of KCl is obtained, what is the percent yield?

Answer 1

Answer:

65.6%

Explanation:

Let's consider the following reaction.

3 Cl₂ + 6 KOH → KClO₃ + 5 KCl + 3 H₂O

We can establish the following relations.

• The molar mass of Cl₂ is 70.9g/mol.
• The molar ratio of Cl₂ to KCl is 3:5.
• The molar mass of KCl is 74.6 g/mol.

The theoretical yield of KCl from 30.0 g of Cl₂ is:

$30.0gCl_{2}.\frac{1molCl_{2}}{70.9gCl_{2}} .\frac{5molKCl}{3molCl_{2}} .\frac{74.6gKCl}{1molKCl} =52.6gKCl$

The real yield of KCl is 34.5 g. The percent yield is:

(34.5 g / 52.6 g) × 100% = 65.6%