Question

245 g pure KCIO
decomposes to give 48 gº,
What is the % yield of
reaction?​

Answer 1

% yield of reaction = 100%

Explanation:

Given: 245 gm of KClO3 on heating yielded 48 gm O2.

Find: % yield of reaction.

Solution:

2KClO3(s) → 2KCl(s) + 3O2(g)

2 mol of KClO3 = 3 mol of O2

1 mol of KClO3 = 3/2 mol of O2 = 1.5 mol of O2

If the reaction has 100% yield, every two moles of potassium chlorate will produce three moles of oxygen gas.

Given that the reaction produces 48g of oxygen gas. Let's see how many moles should be produced.

48 * 1 mole O2 / 32g = 1.5 moles O2

How many moles of KClO3 will we need if the reaction has 100% yield?

1.5 * 2 moles KClO3 / 3 moles O2 = 1 mole of KClO3

% Yield of the reaction = actual yield / theoretical yield * 100

% yield = 48 / 48 * 100 = 100%

Answer 2

% of yield = 100%

hope this will help you