# Answer to Question #198971 in Chemistry for Hsn

Question #198971

2KCLO3→ 2KCl + 3O2(g)

1.5 mole of KCLO3 was decompose during an experiment in laboratory 40 dm3

of O2 was produced find out

percentage yield of O﻿2.

1
Expert's answer
2021-05-31T02:08:37-0400

Solution:

Balanced chemical equation:

2KClO3(aq) → 2KCl(aq) + 3O2(aq)

According to the equation above: n(KClO3)/2 = n(O2)/3

Therefore,

n(O2) = 3 × n(KClO3) / 2 = (3 × 1.5 mol) / 2 = 2.25 mol

At Standard Temperature and Pressure (STP), one mole of any gas occupies a volume of 22.4 L.

Thus 2.25 mol of O2 occupies:

(2.25 mol) × (22.4 L / 1 mol) = 50.4 L O2

50.4 L O2 - theoretical yield

40 dm3 O2 = 40 L O2 - practical yield

Therefore,

%O2 = (40 L / 50.4 L) × 100% = 79.365% = 79.37%

%O2 = 79.37%

Answer: The percentage yield of O﻿2 is 79.37%

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

#### Comments

No comments. Be the first!

### Ask Your question

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS