Question

find the vol of co2 at ntp obtained on heating 10 gm of 90%pure limestone
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Answer 1

Answer:

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

Answer 2

CaCO3 → CaO + CO2
100g → 22.4
10g 100 % pure → 2.24
In case of 90% purity volume(x)
x = 90 x 2.24/100 = 2.016 L