TS EAMCET 4 May 2018 Shift 1 Solved Paper

Molar mass of CaCO3 is 100g∕mol.
10g of 100% pure limestone =10g of CaCO3
So, 10g of 90% pure limestone =9g of CaCO3
This implies that,
9g of CaCO3=0.09‌moleof CaCO3
The chemical reaction occurs during the heating of CaCO3 is written as,

CaCO3

1‌mol

Δ

––––––🢖

CaO+

CO2(g)

1‌mol

By stoichiometry,
1 mole of CaCO3 yields 1 mole of CO2(g) or 22.4L of CO2
The volume of CO2 contained in 0.09‌mole of CaCO3 iscalculated as,
0.09 mole of CaCO3=0.09×22.4L‌of‌CO2
=2.016L of CO2