badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 11

                        

15 mL of 0.5 M H 2 O 2 reacts with 10 mL of 0.5 M KMnO 4 in excess of sulphuric acid, according to reaction 2KMnO 4 + 3H 2 SO 4 + 5H 2 O 2 → K 2 SO 4 + 2MnSO 4 + 8H 2 O + 5O 2 The volume of O 2 produced at STP is

3 years ago

Answers : (1)

Shailendra Kumar Sharma
188 Points
							

2KMnO4 + 3H2SO4 + 5H2O2→ K2SO4 + 2MnSO4 + 8H2O + 5O2

15 ml of .5 M H2O2 i.e. 7.5 mmol of total H2O2 we have 

10 ml of .5 KmnO4 i.e. total of 5 mmol of KMnO4 we have 

Now stoichiometry, says we need 5 moles of H2O2 and 2 moles of KMnO4  which clears that H2O2 is limiting reagent. 
and other thing is with 5 mol of H2O2 gives 5 mole of O2 so 7.5 mmol of H2O2 will give 7.5mmol of O2 = 7.5 *22.4 mL =168mL

3 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details