Question

In a decomposition process, 2.40g of impure CaCO3 was decomposed at 18.0 C and a total pressure of 960.0 torr. The volume of the CO2 obtained, over water, was 2.00L. suppose the water vapor pressure is 16.00 torr at 22.0C. find percent by mass of the pure CaCO3.

Answer 1

CaCO3(S) ---> CaO(s) + CO2(g)

no of mol of CO2 gas = PV/RT

    P = partial pressure of gas = 960-16 = 944 torr

    V = 2 L

    R = 0.0821 l.atm.k-1.mol-1

    T = 22+273.15 = 295.15 k

nCO2 = (944/760)*2/(0.0821*295.15)

       = 0.1025 mol

SO that,

no of mol of CaCo3 present in sample = 0.104 mol

mass of CaCo3 = 0.1025*100

               = 10.25 g

%by mass of the pure CaCO3 = 10.25/2.4*100 = 427%