Question

The volume in litres of co2liberated at ntp when 10 gram of 90% pure lime stone is heated completely.

Answer 1

Explanation:

mass of CaCO3 (limestone) = 100 g. It liberates 22.4 L of CO2. So, 10 g of 90% CaCO3 can liberate (10/100)*(90/100)*22.4 L = 2.016 L of CO2 .

Answer 2

Explanation:

Limestone is essentially CaCO3; molar mass of CaCO3 = 100 g/mol 

10 gm of 100% pure would be 10 gm of CaCO3 

90% pure means you have only 9 gms of CaCO3 in the 10gm sample 

9 gms/100gm/mol = 0.09 moles of CaCO3 

When heated 1 mole of CaCO3 yields 1 moles of CO2 

0.09 moles of CO3 will yield 0.09 moles of CO2 

1 mole of CO2 = 22.4 liters 

22.4 liters/mol x 0.09 moles = 2.016 liters