Question

11 The volume of CO, liberated at 2 atm and 273°C by the thermal decomposition of 200g of 50% CaCO3 sample is show me with solutions ​

Answer 1

Answer:

Correct option is

A

2.016

CaCO

3

⇌CaO+CO

2

One mole of calcium carbonate on decomposition gives one mole of carbon dioxide.

The molecular weight of calcium carbonate is 100 g/mol.

10 g calcium carbonate corresponds to 0.1 mol. It on decomposition will give 0.1 mol of carbon dioxide. But limestone is only 90% pure. Hence, 0.09 mol of carbon dioxide will be obtained.

1 mole of carbon dioxide occupies 22.4 L at STP.

Hence, 0.09 mol of carbon dioxide will occupy 0.09×22.4=2.016 L.