wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Cl2+2KOHKCl+KClO+H2O
3KClO2KCl+KClO3
4KClO3KCl+3KClO4

2 moles of chlorine gas reacted with an excess of potassium hydroxide and 50 g of potassium perchlorate was obtained. Calculate the percentage yield of perchlorate in the reaction.
Molar mass of potassium perchlorate = 138.5 g/mol

A
72 %
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
60 %
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
90 %
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
49 %
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 72 %
The complete balanced equation is:
12Cl2+24KOH21KCl+3KClO4+12H2O

12 moles of chlorine gives 3 moles of perchlorate.
2 moles of chlorine gives 0.5 moles of perchlorate.
Theoretical yield of perchlorate = 0.5 mole = 69.25 g
Experimental yield of perchlorate = 50 g
Percentage yield of perchlorate = experimental yieldtheoretical yield×100 = 50×10069.25=72%

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon