Question

6. In an experiment, 0.5 g of CaCO3 on heating gave
0.28 g of CaO and 112 mL of CO at N.T.P. Show that it
illustrates the law of conservation of mass. (22.4 litres
of CO2 at N.T.P. weigh 44 g)

Answer 1

Given:

The amount of CaCO3 taken = 0.5 gm

The amount of CaO formed = 0.28 gm

The volume of CO2 formed = 112 ml

At NTP, 22.4 L (i.e., 1 mole) of CO2 = 44 gm

To Prove:

This experiment illustrates the law of conservation of mass.

Calculation:

- The thermal decomposition of caCO3 is given as:

CaCO3 → CaO + CO2

- The volume of CO2 formed = 112 ml = 0.112 L

- The mass of 22.4 L CO2 at NTP = 44 gm

⇒ The mass of 0.112 L CO2 at NTP = (44/22.4) × 0.112

⇒ The mass of 0.112 L CO2 at NTP = 0.22 gm

- The total mass of products = Mass of CaO + Mass of CO2

⇒ The total mass of products = 0.28 + 0.22 = 0.5 gm

- Since the mass of reactants = the mass of products, i.e., 0.5 gm, so this experiment illustrates the law of conservation of mass.

- Hence, Proved.