Question

calculate amount of CO2 liberated on decomposition of 50g of limestone​

Answer 1

Answer:

CaCO

3

⟶CaO+CO

2

Moles of CO

2

=

44

11

=0.25

Moles of CaCO

3

=0.25 (stoichiometry ratio 1:1)

Mass =100×0.25=25 gm of pure CaCO

3

%purity=

50

25

×100=50%