Question

to get 5.6 litres of Co2 at STP the weight of caco3 to be decomposed I NEED SOLUTION​

Answer 1

Answer:

Correct option is

25 g

The reaction is CaCO

3

→CaO+CO

2

.

One mole of calcium carbonate forms one mole of carbon dioxide.

At STP, 5.6 L of carbon dioxide corresponds to 0.25 moles.

They will be obtained from 0.25 moles of calcium carbonate.

Molar mass of CaCO

3

=100g

0.25 moles of calcium carbonate (molecular weight 100 g/mole) corresponds to 25 g.