Question

calculate the mass of caco3 required to liberate 10L of co2 at stp?

Answer 1

CaCO3 ------> CaO + CO2

here you can see that

100g CaCO3 decomposed to form 22.4 L gas of CO2 at STP

so,
1L CO2 gas form by 100/22.4 gram decomposed of CaCO3

so, 10L CO2 gas form by 1000/22.4 gram decomposed of CaCO3 .

e.g mass of CaCO3 required = 1000/22.4 gram = 44.64 gram

Answer 2

The required reaction is,

CaCO3 =>CaO + CO2

In STP 44g CO2 is 22.4 litre CO2

So,

22.4 lit CO2 can be required from 100g CaCO2

10 lit CO2 can be required from 100/22.4 × 10g CaCO3

= 1000/22.4

= 44.64g

The answer should be 44.64g CaCO3

Hope it helps