Question

1.25 g of sample of limestone on heating gives
0.44 g carbon dioxide. The percentage purity of
Caco3 in limestone is
(1) 75%
(2) 85%
(3) 90%
(4) 80%​

Answer 1

Answer:

Option(2) I think ! 85%

Answer 2

Answer:

64.8%

Explanation:

The reaction which will take place when limestone is heated will be -

CaCO3 → CaO + CO2

The balanced chemical equation states that when  one mole of calcium carbonate undergoes thermal decomposition at  100

%  yield,  then one mole of carbon dioxide is produced.

Therefore, - 1.25g  limestone gives 0.44 CO2 -

= 1.25-0.44

= 0.81

Thus, by law of conservation of mass -

Mass of CaCO3/mass of Ca(OH)2×100]%

= 0.81/1.25 × 100 %

= 64.8%

Thus, the percentage purity of  Caco3 in limestone is 64.8%.