Question

In a decomposition process, 2.4 g of impure CaCO3 (CaCO3 mixed with other substances that remain as residue) was decomposed at 18 degree C and a total pressure of 960 torr. The volume of the CO2 obtained, over water, was 2.0L. Suppose the Water Vapor Pressure is 16.0 torr at 22.0 degree C. Find the 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 torr

    V = 2 L

    R = 0.0821 l.atm.k-1.mol-1

    T = 22+273.15 = 295.15 k

nCO2 = (960/760)*2/(0.0821*295.15)

       = 0.1042 mol

SO that,

no of mol of CaCo3 present in sample = 0.1042 mol

mass of CaCo3 = 0.1042*100

               = 10.42 g

%by mass of the pure CaCO3 = 10.42/2.4*100 = 434.17%