Question

calculate the volume in litre of co2 liberated at ntp when 45 gram of 90% pure limestone heated completely??​

Answer 1

Explanation:

CaCO

3

Δ

CaO+CO

2

↑

The molar mass of CaCO

3

=100 g/mol.

10 g of 90% pure lime

=

100g/mol

10g

×

100

90

=0.09 moles CaCO

3

0.09 moles CaCO

3

on heating gives 0.09 moles CO

2

.

At STP. 1 mole CO

2

= 22.4 L.

At STP. 0.09 mole CO

2

=0.09×22.4=2.016 L.

Hence, the volume of CO

2

liberated at STP when 10 g of 90% pure lime is heated completely is 2.016 L.