The correct option is B 2.016 L
CaCO3→CaO+CO2
As we are given that CaCO3 is 90% pure
So, 9 grams of limestone is present in 10 grams of the taken sample.
Number of moles of CaCO3
=9100 mole
We know that:
Moles of CaCO3=moles of CO2=0.09 mole
volume of one mole of gas at STP =22.4 L
So volume of CO2
=0.09×22.4=2.016 L