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: 12th pass
        
The decomposition of a certain mass of CaCO3 gave 11.2 litre of CO2 gas at STP. The mass of KOH required to completely neutralise the gas is: 
a.  56 g
b.  28 g
c.  42 g
d.  20 g
 please explain in detail...
6 months ago

Answers : (1)

Arun
21031 Points
							
At STP 22.7 L of an ideal gas contains 1 mol of its component, hence in 11.2 L of CO2 the moles of CO2 present = (1/22.7) x 11.2 = 0.493 mol

​Reaction of KOH with CO2​ :      2KOH + CO2 ----> K2CO3 + H2O

​Hence, there is 2 mol of KOH required for neutralising 1 mol of CO2 , hence for 0.493 mol of CO2 , moles of KOH required will be = 2 x 0.493 = 0.986 mol

Mass of KOH required = 0.986 mol x 56 g mol-1 = 55.2 g
6 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

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


Course Features

  • 141 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