0

Guest

Courses New

The decomposition of a certain mass of CaCO 3 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...

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...

Grade:12th pass

1 Answers

Arun
25750 Points
5 years ago
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

Think You Can Provide A Better Answer ?

Other Related Questions on Physical Chemistry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More